LIBRARY OF CONGRESS. 



C^qt^Ste&igp $o.- 

Shelf ,W&i- 

UNITED STATES OF AMERICA. 



KEY MD SUPPLEMENT 



Elementary Mechanics. 



BY 

BE VOLSON WOOD, 






' V- J I 

/i U4 t vv 

^ Of v./,. 



JOHN WILEY & SONS, 
New York. 



^ 






Copyright, 1882, 
By DE VOLSON WOOD. 



NOTICE. 



This work contains not only solutions of the exam- 
ples and answers to the Exercises of the Elementary 
Mechanics of the author ; but also additions to the text, 
and other new matter intended to interest the general 
student of mechanics, and be of service to teachers of 
the science in connection with any other text book. 
Valuable assistance lias been rendered in correcting 
proofs by Professor IT. A. Howe, of the University of 
Denver. 

DeVolson Wood. 



KEY AND SUPPLEMENT 



ELEMENTARY MECHANICS. 



Page 1, Article 1. — Motion is a change of position. 
Motion is determined by the relative position of bodies 
at different times. If bodies retain the same relative 
position during successive times, they are said to have 
no motion in reference to each other; in other words, 
they are said to be at rest in reference to each other. 
All bodies of which we have any knowledge are in 
motion ; hence all motion is, so far as we know, rel- 
ative. Absolute motion implies reference to a point 
absolutely at rest, but as no such point is known, such 
motion -has only an ideal existence. 

Page 2, Art. 6. — "No definition of space will give a bet- 
ter idea than that obtained by experience. Metaphysi- 
cians have indulged in speculations in regard to its na- 
ture, but they are able to assert with certainty only 
that it has the property of extension. Descartes taught 
that the properties of extension, known as length, 
breadth, and thickness, were solely properties of mat- 
ter, and hence when a body was removed no space 
remained in the place formerly occupied by it. So 
far as we know, no space exists which is perfectly de- 
1 



KEY AXD SUPPLEMENT 

void of matter. Perfectly void space is an ideality ; 
still modem philosophy distinguishes between the 
thing contained, and that which contains it-l^tween 
matter and the place occupied by matter. It abstract, 
(so to speak) space from matter, and, m a measure, 
Liter from space. It seems impossible to conceive 
of matter not occupying space, but it is not d.ftcult to 
conceive of a given quantity of .natter as occupying 
a very small or a very large space. We are able to con- 
eider matter in the abstract without considering the 
dimensions of the space it occupies ; and also we may 
consider space in the abstract as not including matter 
The latter is called absolute space; it is conceived 
as remaining always similar to itself and immovable 

Time is duration. We gain a knowledge of it by 
the order of events. Every event has its place in 
time and space, and by means of memory we gain a 
knowledge of the order in which events occur. \\ itli- 
out memory we would gain by experience no knowl- 
edge of time. Sir Isaac Newton considered mathe- 
matical time as flowing at a uniform rate, unaffected 
by the motions of material things. This idea induced 
him to call his new calculus fluxions. 

Rate refers to some unit as a standard. Thus, to 
illustrate, rate of interest is a certain amount of money 
paid for the use of one dollar; passenger rates refer 
to the amount paid for one passenger ; rate of shipping 
per ton is the amount paid for carrying one ton ; rate 
of motion is the space passed over in one second, om 
minute, one hour, one day, or one of any other umt. 
The term velocity is simply the equivalent of the rati 
of motion. Angular velocity is rate of angular motion 



TO ELEMENTARY MECHANICS. 3 

(p. 4, Art. 12). Acceleration is the rate of change of 
velocity, being the amount of change in the velocity 
for one second, or one of any other unit (p. 10, Art. 
22). Mechanical power is the rate of doing work (p. 
55, Art. 99). Rates may involve two units. Thus, 
rate per ton-mile implies a certain amount paid for- 
one ton for one mile ; passenger rates are often an 
amount for one person for one mile ; mechanical power, 
or rate of doing work, is the amount of work done in 
one foot for one second, or by one pound for one sec- 
ond, etc. Rate is a thing used for measuring quantities, 
as a yard-stick is used for measuring cloth, a chain 
for measuring land, the pound for weighing groceries, 
ton for measuring merchandise, etc. 
Page 4, Art. 12. — The definition here given for rotary 
motion is applicable to the 
case where the motion is in ^ _^~^/D 




a curved path not circular, q , 

as CD. But the analysis 

given in the text is not applicable to this case. 

Page 6, Art. 14. — Just after Fig. 5, for If two ve- 
locities, etc., read If two concurrent simultaneous 
velocities, etc. 

Page 8, Art. 20. — Speaking of the' rotation of the 
moon, suggests an interesting question in practical 
mechanics. If the wheel 
B rolls around on the cir- 
cumference of an equal J 
wheel A, will the former ' 
turn once or twice on its 
own axis? 

Mark the point a which, initially, is in contact with 




KEY AND SUPPLEMENT 



the wheel A, and roll B half around A ; it will be 
observed that the marked point will then be at the 
left of the centre of B, as it was at the start. Con- 
tinuing the rolling, the point a will again be at the 
left of the centre when B has gone completely around 
A ; hence it is sometimes asserted that the wheel B 
has turned twice on its axis. But we wish to show 
that it has turned but once on its own axis, and the 
whole wheel has been rotated once about the axis of 
the wheel A. Let the axis of the wheels be at right 
angles with each other, then 
it will be evident, from 
mere inspection, that when 
B has turned once on its 
axis it will have gone once ' 
around A. Thus B will have 
gone once around the axis of 
A, and once about its own 
axis. Xext, incline the axis 
of B upward, so as to ap- 
proach a parallelism to A, 
and the same result will be 
seen from mere inspection, 
and it will continue to remain evident as it be- 
comes nearer and nearer parallel, and when they 
become actually parallel, the same condition will 
hold true. Hence, in the former figure, the wheel 
B will turn but once on its own axis in rolling once 
around A. The same result may be shown in an- 
other way. Let a block be placed at a facing a mark 
on the axis of B, and conceive this axis to be rigidly 
connected with the axis of A while the wheel B is 




TO ELEMENTARY MECHANICS. 5 

free to turn on its own axis. In this way the axis of 
B will be carried bodily about A. When B has rolled 
half around A, it will be found that the block will 
face the same direction in space — say towards the 
east — but that it will not face the mark on the axis, 
for the mark will be on the opposite side of the axis. 
Continuing the rotation, it will be found that the 
block will face the mark only once at each reyolution 
about A. 

Similarly, the moon turns but once on its own axis 
in one reyolution about the earth, but the rotation 
about the two centres are not exactly coincident ; for 
it is found by observation, that in some parts of the 
orbit more of the surface of the moon is seen on the 
eastern (or western) side than in other parts of the 
orbit ; thus showing that the rotation about the earth 
is sometimes faster, and at other times slower than the 
rotation of the moon about its own axis. This phe- 
nomenon is called Libration. 

exeecises. 
Page 9. 

1. 4-J- miles. 

2. The former. 

3. 66 feet per second. 

4. 17 feet ; 17n feet. 

■ 4U ° ■ 60 x 60 ~ T1 seconCls - 

■, 200 x 2* 200 x 360 1Qnn 

6. — — = 6§ 7t m arc ; or — = 1200 

degrees. 



7. V3" + 2 J ~ Vl3 =2 3-605 + miles per hour. 



KEF AND SUPPLEMENT 



Page 10. 



i/i 



[~\ = £^3961 = 20-98 ft. per sec- 



ond. 
9. 0-00091. 

10. See Article 14. v = Vo 2 + 10 J + 100 cos 60° 
= Vl25 + 50 = V1T5 = 13-22 + feet ; hence 
the distance between them in two seconds will 
be 2 x 13-22 + = 26-44 + feet. 
Page 10, Art. 22. — Observe that acceleration is not the 
rate of change of motion, but the rate of change of 
the rate of motion. It is the rate of change of a rate. 
The rate of change is usually measured in the same 
units as the rate of motion. If one is in feet per sec- 
ond, the other is also. It is possible to conceive of 
mixed units. Thus in the case of falling bodies, the 
velocity at the end of the first second is 16 T V feet, and 
the acceleration is 643^ yards per minute. 

Strive to get a clear conception of the meaning of 
acceleration and of its measure. It is one of the ele- 
ments of the absolute measure of force. 
Page 13, Art. 26. — The expression " The locus of these 
points will be a parabola," means 
that if any number of points in 
the path be determined in the 
same manner, they will all be 
in the arc of a parabola. 

A parabola is a curve which 
may be cut from a right cone 
by a plane parallel to one of its 
elements. (See Author's Coordinate Geometry.) 




TO ELEMENTARY MECHANICS. 7 

EXAMPLES. 

1. Space 250 feet ; velocity 50 feet per second. 
Page 14. 

2./ =25 feet; s = 12J- feet. 

3. During 4 seconds the space will be .5 = -|- of 32 

x 4 2 (Art. 14), and during 3 seconds the space 
will be \ of 32 x 3 2 ; hence during the 4th second 
the space will be \ x 32 (4 2 - 3 2 ) = 112 feet. 

4. By means of equation (1), Art. 25, find 6J feet 

per second. 

5. t= V12-5 = 3-533 4- seconds. ^ 

6. /= (20 4- 120) 3-23 = 0-54| feet per second. 

7. 321 _i_ 3.28 = 9-8 metres per second. 

Page 15. — Matter and force are two grand realities of 
the external world, and of these we know nothing 
directly. Our knowledge of the former is confined 
to its properties^ and of the other to its laws of action. 
But we have no reason to believe that one exists in- 
dependently of the other. In our earlier experiences 
matter is conceived to be hard, gross, and unyielding ; 
but later we find that it is yielding, and that many 
solids, as iron and lead, may be changed to liquids by 
heat, and that liquids may be changed to gases — so 
that matter is proved to be more or less viscid or at- 
tenuated. Solids are porous. Changes of form are 
effected by forces, so that some metaphysicians have 
reasoned that there may possibly be no gross matter, 
but, instead thereof, those things which we consider 
as bodies are only aggregations of forces. On the 
other hand, all investigations in mechanics proceed 
on the hypothesis that matter is in no sense a force, 



g KEY AND SUPPLEMENT 

or an aggregation of forces, but that it is something 
distinct from force, something upon which force nets. 
In the case of attraction, the matter in two bodies 
may remain constant, while the force exerted by each 
upon the other will depend upon the distance be- 
tween them. It is true, however, that we gain a 
knowledge of matter only through the action of forces. 
Every avenue to the mind through the senses is an 
agent for transmitting the result of the action of cer- 
tain forces, and the very act of transmission brings 
into piay certain forces. 
Page 16, Art. 28.— Mathematics applied to the laws of 
physical science enables ns to determine magnitudes 
which far transcend the powers of accurate measure- 
ment or even of conception. 

Sir Wm. Thompson gives four methods for ascer- 
taining the mean distance between molecules. 
Optical dynamics. 
Contact electricity of metals. 
Capillary attraction. 
Kinetic theory of gases. 

" Optical dynamics leaves no alternative but to admit 
that the diameter of a molecule, or the distance from 
the centre of a molecule to the centre of a contiguous 
molecule in glass, water, or any other of our transpa- 
rent liquids and solids, exceeds one ten-thousandth of 
the wave length of light, or a two-hundred-millionth 
of a centimetre" ( a 6oo oW o<n r of a metre). 

"However difficult it may be even to imagine what 
kind of thing the molecule is, we may regard it as an 
established truth of science that a gas consists of mov- 
ing molecules disturbed from rectilineal paths and 



TO ELEMENTARY MECHANICS. 9 

constant velocities by collisions or mutual influences, 
so rare that the mean length of proximately rectilineal' 
portions of the path of each molecule is many times 
greater than the average distance from the centre of 
each molecule to the centre of the molecule nearest 
it at any time. If for a moment we suppose the mole- 
cules to be hard elastic globes all of one size, influenc- 
ing one another only through actual contact, we have 
for each molecule simply a zigzag path composed of 
rectilineal portions, with abrupt changes of direction.'' 
" If the particles were hard elastic globes, the aver- 
age time from collision to collision would be inversely 
as the average velocity of the particle. But Max- 
well's experiments on the variation of the viscosities 
of gases with change of temperature prove that the 
mean time from collision to collision is independent 
of the velocity, if we give the name collision to those 
mutual actions only which produce something more 
than a certain specified degree of deflection of the 
line of motion. This law could be fulfilled by soft 
elastic particles (globular or not globular), but not by 
hard elastic globes." " By Joule, Maxwell, and Clau- 
sius we know that the average velocity of the mole- 
cules of oxygen, or nitrogen, or common air, at or- 
dinary atmospheric temperature and pressure, is about 
50,000 centimetres per second (500 metres per sec- 
ond, or about 1,600 feet per second), and the average 
time from collision to collision a five-thousand- 
millionth of a second (jowfohit)' Hence the aver- 
age length of path of each molecule between collis- 
ions is about 1QQ 1 Q0Q of a centimetre " ( T „ 7o 1 ooou of a 
metre). 



10 KEY AND SUPPLEMENT 

" The experiments of Cagniard de la Tour, Faraday, 
Regnault, and Andrews on the condensation of gases 
do not allow us to believe that any of the ordinary 
gases could be made forty thousand times denser than 
at ordinary atmospheric pressure and temperature 
without reducing the whole volume to something less 
than the sum of the volumes of the gaseous molecules, 
as now defined. 

" Hence, according to Clausius, the average length 
of path from collision to collision cannot be more 
than five thousand times the diameter of the gaseous 
molecule ; and the number of molecules in unit of 
volume cannot exceed 25,000,000 divided by the vol- 
ume of a globe whose radius is that average length of 
path. Taking now the estimated T o oWo" 0i? a centime- 
tre for the average length of path from collision to 
collision we conclude that the diameter of the gaseous 
molecules cannot be less than g o o uV o~ o o °^ a centime- 
tre Goimo oinnroo" 0i? a metre) ; nor the number of 
molecules in a cubic centimetre of the gas (at ordi- 
nary density) greater than 6,000,000,000,000,000,000,-' 
000." 

"The densities of known liquids and solids are 
from live hundred to sixteen thousand times that 
of atmospheric air at ordinary pressure and tempera- 
ture : and, therefore, the number of molecules in a 
cubic centimetre may be from 3 x 10 24 to 10 2 G (that is, 
from three million million million million, to a 
hundred million million million million). From 
this the distance from center to nearest center in 
solids and liquids may be estimated at from y^o-ffVinmF 
{ .° iTwwiwuo-o 0T " a centimetre ( 1Tjnni \ t -. , to 



TO ELEMENTARY MECHANICS. IX 

innro looooo of a metre). The four lines of argu- 
ment lead all to substantially the same estimate of 
the dimensions of molecular structure. Jointly they 
establish with what we cannot but regard as a very 
high degree of probability the conclusion that, in 
any ordinary liquid, transparent, solid, or seemingly 
opaque solid, the mean distance between the centers 
of contiguous molecules is less than the hundred- 
millionth, and greater than the two thousand-millionth 
of a centimetre. To form some conception of the 
coarse-grainedness indicated by this conclusion, im- 
agine a rain drop, or a globe of glass as large as a pea, 
to be magnified up to the size of the earth, each con- 
stituent molecule being magnified in the same pro- 
portion. The magnified structure would be coarser- 
grained than a heap of small shot, but probably less 
coarse-grained than a heap of cricket-balls." (Ex- 
tracts from a paper by Prof. Sir ¥m. Thomson on 
the size of Atoms, Am. Jour, of Science and Art. 
1870, vol. ii., pp. 38-45.) 

Mr. N. D. C. Hodges, in an article on the size of 
molecules (Phil. Mag. and Jour, of Science, 1879, 
vol. ii., p. 74), says : " If we consider unit mass of 
water, the expenditure on it of an amount of energy 
equivalent to 636.7 units of heat will convert it from 
water at zero into steam at 100°. I am going to con- 
sider this conversion into steam as a breaking-up of 
the water into a large number of small parts, the total 
surface of which will be much greater than that of 
the water originally. To increase the surface of a 
quantity of water by one square centimetre requires 
the use of .000825 metre gramme of work. The total 



12 KEY AND SUPPLEMENT 

superficial area of all the parts, supposing thcin spher- 
ical, will be 4 re rN, the number of parts being X. 
The work done in dividing the water will be 4 n r % 
N x .000825. For the volume of all the parts we 
have | 7T r*N. This volume is, in accordance with 
the requirements of the kinetic theory of gases, about 
3,000 of the total volume of the water. The volume 
of the steam is 1,752 times the original unit volume 
of water. Hence — 

A^/' 3 iT3000 = 1752 

4 7t r 2 1ST. 000825 = G3G.7423. 

One unit of heat equals 423 units of work (in French 
units); solving these equations for /' and iV, we get 
r = .000000005 centimetre (or diameter = .00000001 
centimetre = TWuoohowows metre), a quantity closely 
corresponding with the previous results of Sir Win. 
Thomson, Maxwell, and others ; and N equals 9000 
(million) 3 , or for the number in one cubic centime- 
tre 5 to 6 (million) 3 ." 

The extreme tenuity of a gas is further shown by 
the following extract taken from the Beiblatter zu 
den Annalen der Phy&ik und Che?nt€, 1879, No. 2, 
]>. 59. "At 0°C. and 760mm pressure a cubic cen- 
timetre (.0G1 cubic inch) holds nearly one hundred 
trillions of gas molecules. Under these conditions 
the molecules themselves fill nearly the T f n of the 
space occupied by the gas. The absolute weight of a 

15 

hydrogen molecule is represented by — — g, ((/ in me- 
tres)." 
Mr. G. J. Stoney, in an article on Polarization on 



TO ELEMENTARY MECHANICS. 13 

Stress in Gases {Phil. Mag. and Jour, of Science, 1878, 
vol. ii., p. 407), says that " the number of molecules in 
a cubic millimetre of atmospheric air is about 10 18 
( = 1,000,000,000,000,000,000),* and that the average 
distance between them is about y-gij-W o o o o o °^ a m etre. 
The average striking distance (i. e. the average length 
of path between encounters) of the molecules is about 
T^ oo 1 ooft TT °f a metre. The average velocity at ordi- 
nary temperature may be taken as 500 metres per sec- 
ond (1,600 feet per second), and the molecules meet 
with so many encounters, that the direction of the 
path of each is changed somewhere about 10,000,000,- 
000 times every second." In one movement the par- 
ticle travels 20 oirWtro- °^ a merre ) or 20-0017 °f a milli- 
metre ;t and it makes this movement in 10 ooo1)ooo inr 
of a second. jSTow the wave length of a chemical 
ray is about -g ¥ Vo °f a millimetre, hence we find that 
the molecule of air travels through a distance which 
is one-fourth as long as the length of this particular 
wave in this fraction of a second. % 

According to Pouillet, the mechanical energy of a 
cubic mile of sun light at the earth equals 12,050 ft. lbs. 

* Clausius's limit is 6,000 times this amount. 

f Clausius's estimate is one-lialf this value. 

\ Mr. E. H. Cook, in an article on the Existence of the Lumi- 
niferous Ether {Phil. Mag. and Jour, of Science, 1879, vol. I, p. 
235), after quoting the above figures from Mr. Stonev's paper, 
adds : "Then in one moment the particle travels 2000000 of a 
metre in tuouooottoutt of a second;" also in his deductions he 
adds : "hence we find that the molecule of air travels through a 
distance which is more than twice as long as the length of this 
particular wave in this fraction of a second." The reader can 
easily see that this deduction is erroneous. 



14 KEY AND SUPPLEMENT 

{Phil. Mag., 1855, vol. ix., p. 39) ; accepting which 
Sir Wm. Thomson calculated the weight in pounds 
of a cubic foot of the ether of space (that which is 
instrumental in transmitting light, and called ether) 
by the formula : 

83r/ 



W = 



r 



where g — 32 \ (acceleration per second of gravity), 
J"— the velocity of light per second, being about 
192,000 miles per second, and n = -fa, being the ratio 
of the greatest velocity of a rotating particle to the 
velocity of light. Herr Grlan asserts that n is not 
constant, that he found n — -^ in one case, and 53 1 (TI 
in another. {Am. Jour, of Arts and Sciences, 1ST9, 
vol. xviii., 404. Annalen der Physik und Cheinie, 
No. S, 1S79, p. 5S4.) Assuming n — ^, we find that 
a cubic foot would weigh about 

w 83 x 321 x 502 



(52S0 x 192000) 3 

i- fi n i 

=ioTo lb - nearl .y; 



and for the weight of a cubic mile pound. The 

weight of a volume of the size of the earth would be 
about 240 pounds. Admitting this result, it follows 
that a sphere equal in diameter to the diameter of the 
earth's orbit (or say 190,000,000 miles) will contain 
an amount of ethereal matter nearly equal to j-^ 
of that of the mass of the earth. 

Probable tension of tJie ether of space. The above 



TO ELEMENTARY MECHANICS. 15 

results combined with one or two other plausible 
assumptions enable us to find a probable value of 
the tension of the light ether. As stated above, 
the velocity of the particles of air producing a 
pressure of 15 lbs. per square inch, is about 1,600 
feet per second, which is about 50 per cent, more 
than the velocity of sound in air. Assuming now 
that the normal velocity of the ether particles is 
somewhat more than the velocity of light — or, more 
definitely, that it is 195,000 miles per second, and 
that the tension is directly as the masses, and also as 
the square of the velocities of the particles ; also that 

the weight of a cubic mile of the ether is — — lb., as 

a 10° 

found above, and observing that 100 cubic inches 
of ordinary air weigh 31 grains, and 7,000 grains 
make a pound, we have for the pressure P in pounds 
of the ether upon a squa"re inch 

1 

n _ r / 195000 x 5280 X 2 10 9 

- ° X \ 1600 J X 31 1728 

loo x 7ooo x(o ^ 0) 

= 0-00000185 = ^? lb. per inch of section. 
10 6 x 

A Mr. Preston, an English writer, in his work on the 
Physics of Pike? 1 , estimates, or rather assumes, 500 
tons per square inch as a probable inferior limit of 
the pressure (p. 18), and, with this as a basis, he finds 
the weight of a cubic mile of ether to be about 220 lbs. 
(p. 120). But as he has used 56-5 grains for the 
weight of a cubic foot of air when it is nearly ten 



IQ KEY AND SUPPLEMENT 

times this amount, lie should have found for the 
weight of a cubic mile of ether nearly one ton. His 
assumption, however, in regard to pressure is quite 
arbitrary, and does not seem to be well founded. It 
seems improbable that there should be so much mass 
in the ether of space. Even 240 lbs. for a volume 
equal to that of the earth seems a high value when we 
consider the amount that must be displaced by the 
planets while moving about their orbits. 

Temperature of the Sun. The following may be interesting-, 
although it does not fall under the article above referred to. 
" The effective temperature of the sun may he defined as that 
temperature which an incandescent body of the same size placed 
at the same distance ought to have in order to produce the same 
thermal effect if it had the maximum emissive power. If we 
consider the surrounding temperature during the observation to 
have been about 240° Ave obtain . . . for the effective tempera- 
ture in degrees centigrade 9,905.4'. . . I think, then, that I may 
fairly conclude that the temperature of the sun is not very dif- 
ferent from its effective temperature, and that it is not less than 
10,000°, nor much more than 20,000° centigrade." Phil. Mag., 
1879, vol. ii., pp. 548-550. See also Am. Juiir. Sc. and Arts, 
1870, vol. ii., p. 63. 

Sir Win. Thomson calculates the mechanical energy of the 
solar rays falling annually on a square foot of land in latitude 50' 
to equal 530,000,000 foot pounds, or 396 H. P. per yard per day. 
He finds tha*< the heat alone hourly given out by each square 
yard of the solar surface is equivalent to 03,000 horse power, and 
would require then the hourly combustion of 13,500 lbs. of coal. 
Appleton's Cyclopedia, 186S, vol. ix. p. 2'-). 

Page 16, Art. 29. — A better definition is — Force is an 
action between oodies — for this form of the statement 
recognizes the existence of at least two bodies in 
every action. A force never acts upon one body 
without producing an equal opposite action upon an- 



TO ELEMENTARY MECHANICS. 17 

otlier body. We speak of the action of a force upon 
a body because in most cases the second body is so 
large, relatively, that the force produces little or no 
perceptible effect upon it. But according to the law 
of Universal Gravitation each particle in the universe 
is attracted by every other particle with a force which 
depends upon their masses and the distances between 
them ; hence, in a highly refined sense, it may be said 
that every force producing motion involves every 
body in the universe. The entire universe of matter 
is bound together by a something — an action — which 
we call Force. Every phenomenon which we witness 
in the physical world is the result of force , acting 
through space, or during a certain time. 

Force alone is stress. In other words stress is 
force abstracted from time and space. The science 
of Stress is the science of Statics. Stress is always 
measured in pounds or its equivalent. If a force pro- 
duces motion, that part of the phenomenon which is 
abstracted from time and space is stress, so that the 
attractive force between the earth and moon, or be- 
tween the earth and sun, measured in pounds, is 
stress. 

When force is compounded with time or space the 
result is work, or energy, or momentum, as will here- 
after be shown. 

The following are some definitions of force as given 
by different authors : 

La Place says : " The nature of that singular mod- 
ification, by means of which a body is transported 
from one place to another, is now, and alwa} T s will 
be, unknown ; it is denoted by the name of Force. 



18 KEY AND SUPPLEMENT 

We can only ascertain its effects, and the laws of its 
action." (3£ecanique Celeste, p. 1). 

"iforce is an action between bodies, causing or 
tending to cause change in their relative rest or mo- 
tion " (Bankings Applied Mechanics, p. 15). 

u Force is that principle of which, considered sim- 
ply as a mechanical agent, we know but little more 
than that when it is imparted, that is, put into, a 
body, it produces either motion alone ; or strain, with 
or without motion." " What is called overcoming 
inertia, is simply putting in force? (Tnuitwine's 
Engineer } s Pocket Book, p. 415 and p. 447). We 
consider this as a misuse of terms. We cannot safely 
say that force is put into a body. When force acts 
upon a body free to move, energy is put into the 
body, when force and space are involved in the result, 
and time is abstracted (see text, p. 66) ; or momentum 
when the elements involved are force and time, space 
being abstracted (see Chap. V ). Although it is too 
early in the text for a full discussion, we add a re- 
mark for the benefit of those who have some knowl- 
edge of the subject. When a constant force, F, acts 
through a space, s, it does the work represented by 
the product of F and s, or Fs. If the body upon 
which it acts is wholly free, the entire work will be 
stored in the body, and is then called energy, the 
measure of which is \ Mv*; hence Fs —\Mtf. Elimi- 
nating 8 by means of Eq. (2), p. 12, of the text, gives 
Ft = 2fv, which is the measure of the effect of force 
combined with time, and the .second member is the 
measure of the momentum (text p. 78). It would be 
better to say that force and time, or force and space, 



TO ELEMENTARY MECHANICS. 19 

are put into a body than that force alone is put into 
it. " .Nothing but force can resist force." " Matter, 
in itself, cannot resist force " (ib. p. 445). Bj r 're- 
sistance is here understood to be such a condition of 
things as that motion will not result, and, in this 
sense, these statements are correct. 

"Force put into a body " is, properly speaking, put- 
ting it under stress, and the body is said to be strained, 
but no amount of internal stress will produce motion 
of the body. 

" Whatever changes the state of a body or the ele- 
ments of a body, with respect to rest and motion, is 
called force." (Bartlett's Analyt. Mech., p. 17). 

" Force is detined as that which changes or tends 
to change a body's state of rest, or motion, and any 
given force may be measured by the acceleration it 
imparts to a gramme." (Cumming's Theory of Elec- 
tricity, p. 5). 

" Force is whatever changes or tends to change the 
motion of a body by altering either its dimension or 
its magnitude ; and a force acting on a body is meas- 
ured by the momentum it produces in its own direc- 
tion in a unit of time." (Maxwell's Theory of Heat, 
p. 83). 

It will appear that the momentum produced in a 
unit of time is the same as the acceleration, and hence 
the two last definitions are equivalent. We, however, 
deem it advisable to avoid the expression momentum 
produced because it is liable to be confounded with the 
actual momentum of a body, although it is not in- 



20 KEY AND SUPPLEMENT 

tended to even imply the latter. Acceleration is 

specific and correct. 

"Force is matter in motion, nothing more, nothing 
less ; the abstract idea of force without matter is a 
nonentity." (Nystrom On the Force of Falling Bodies 
and Dynamics of Matter ', p. 20). 

The preceding remarks will show that this defini- 
tion contains a misapplication of terms. Matter in 
motion is either Energy or Momentum, according as 
time or space is abstracted in considering the ele- 
ments which enter into the combination. 

" Force is a mere name, but the product of a force 
into tl i e displacement of its point of application has 
an objective existence." 

" Force is the rate at which an agent does work 
per unit of length." 

" The mere rate of transference of energy per unit 
of length of that motion is, in the present state of 
science, very conveniently called force." (Lecture by 
Prof. G. P. Tait, Nature, 1876, vol. xiv. p. 462). 

In regard to these views, we observe that the name 
applied to anything is a mere name. In a certain 
sense it is an ideality, but generally the name stands 
for a reality. In this case, if force is a mere name, 
what is the sense of the remainder of the sentence l 
How can a force have & point of application if the 
force is a mere name? And granting that it may 
have, how can the product have an objective exist- 
ence? 

In regard to the second definition, it is analytically 
correct, but we consider it rather as a deduction than 



TO ELEMENTARY MECHANICS. 21 

as a fundamental definition. It is shown on pages 
52 and 67 that 

or, in terms of the calculus, 

■Fds = d'(iJLf) = \Md{f) = dK. 
From the former we have 

JP _ jMv* 

and from the latter 

* ~~ ds > 

hence, generally, force is the rate of doing work per 
unit of length. But is force merely a rate ? What 
shall be said of force as a stress, where no transfer- 
ence of energy takes place ? That this definition is 
not elementary, but a mere deduction, is not only evi- 
dent, but may be more forcibly shown by means of 
other deductions. Thus, from the former equation 
we have 



hence, space is the rate of doing work per unit of 
force (per pound). 
Or again, 



v = \/\ 



2** 



22 KE Y AND SUPPLEMENT 

lience velocity is the square root of twice the rate of 
doing work. per unit of mass ! 

Or, again, in regard to momentum, page 78, 
^Ft = Mo ; 

hence, force is the rate of producing momentum per 
unit of time. 
Or, again, 

. Mv 

hence, time is the rate of producing momentum per 
unit of force ! 

Now all of these are correct deductions ; but the 
fundamental equations are established on the hypothe- 
sis that all, except velocity, are substantial quantities. 
The value of Fis fundamentally measured in pounds. 
Indirectly it may be measured in a variety of ways as 
shown above, and as will be still further shown in 
Article 86 of the text. 

Page 17, Arts. 31 and 32.-— Terrestrial Gravitation as 
a force causes, or tends to cause, bodies to move to- 
wards the earth, and when a spring balance or other 
weighing machine is interposed to prevent any move- 
ment, the intensity of the impelling force may be. de- 
termined in pounds or an equivalent. 

Page 19, Art. 35.— The fundamental idea of a point of 
application of a force is that of a definite attachment, 
like the attachment of a rope, or chain, or rod of iron, 
to a body ; but it is certain, in regard to the forces of 
nature— as gravity, chemical forces, etc.— that the 
conception is erroneous, for there is no attachment. 



TO ELEMENTARY MECHANICS. 23 

Still it may be conceived that the force acts upon a 
particle which may still be considered as the point of 
application, and thus the old term, with its gross as- 
sociations, is useful in the most refined sense. 
Page 20, Art. 36. — Inertia is a name merely to express 
the fact that matter has not of itself power to put 
itself in motion, or being in motion to bring itself to 
rest, or even to change its rate of motion. Yet some 
writers call Inertia a force, and others,- with little if 
any more propriety, speak of the force of inertia. M. 
Morin, a French physicist, attempted to prove that 
inertia is a force. He took a prism standing on its 
base, and by a sudden pull or push applied to its base 
caused the prism to fall backwards. He argued that 
the falling over of the prism indicated the action of a 
force. A force might have been applied at the top of 
the prism which would have overturned it directly, 
and Morin argued, that when it overturned by a sud- 
den action at the base, there must have been a force 
equivalent to one at the top, and this he called the 
force of inertia. The fact is, any force applied to a 
body, not acting in a line through its centre, tends to 
rotate the body ; and in all cases where the body is 
free, will cause it to actually rotate. If the prism 
referred to, standing on its base, be acted upon by a 
force applied at any point above the base, it will not 
be overturned unless the moment of the force exceeds 
the moment of the weight in reference to the point 
about which it tends to turn. If the force applied at 
the base be so intense as to produce a rotary mo- 
ment exceeding the moment of the weight above re- 
ferred to, it will cause the prism to fall by rotating 



21 KEY AND SUPPLEMENT 

backwards ; but if the force be less intense, it will 
simply cause the body to slide on the plane. 

Again, inertia does not fulfill any of the conditions 
of a force. It is not an action between bodies. It is 
not an action in any sense. It cannot be measured 
by pounds. It is a negation — an entire lack of some- 
thing—a lack of force. We repeat, it is not a force. 

exercises. 
Page 22. 

1. The least force. One object of some of these 
exercises is to enable the student to get a correct 
idea of the relation between force and the result- 
ant motions of bodies by the inductive method. 
The student who has not correct notions of these 
relations will doubtless insist that it requires more 
force to move a large body than a small one; 
and he may go so far as to say that it will take 10 
pounds of pull or push to move a body weighing 10 
pounds. But the fact must be perceived that the 
smallest force (10 lbs. of push, for instance) will just 
as certainly move 100 lbs. of matter free to move in 
the direction of the push, as it will one pound. It will 
not move the former as rapidly as the latter— or, in 
other words, it will not move it as far in the same 
time. In this way we get an idea of the fact that the 
visible effect of a force depends conjointly upon the 
mass moved, and the space through which it is moved 
in a given time. If necessary make an experiment 
by suspending different weights with equally long- 
strings, and pull them sidewise by a string attached 
to the body passing over a pulley — or edge of the 



TO ELEMENTARY MECHANICS. 25 

table — and holding at the suspended end a small 
weight. It will be found that any weight which is 
. sufficient to overcome the friction of the string on the 
pulley — or edge of the table — will pull the heaviest 
weight sidewise. Observe that the experiment is 
simply to show actual movement, and not the amount 
of movement. 

2. The least force. Also the least force would de- 
flect it from its course. This shows that a force will 
have the same effect upon a moving body as upon one 
at rest. 
Page 23. — 3. Because it is opposed by an equal oppo- 
site force. 

4. 100 pounds. A man once pulled a spring bal- 
ance so as to indicate, say, 100 pounds. Another gen- 
tleman asserted that he could pull two balances at- 
tached end to end so that each would indicate 100 
pounds. This he did to the astonishment of the ob- 
servers, but their astonishment ceased when they 
found that it was no more difficult to pull two in that 
way than one. 

5. Yes. The fact that the boat is in motion, does 
not affect the result. Hence we have the principle 
that " action and reaction are equal and opposite." 

6. J^o. To show it (should there be doubt) assume 
that a string passes from each sled to the hand, then 
will the tension on each be less than 10 pounds, but 
on both strings it will be just 10 lbs. Conceive that 
the strings become one back to the first sled — the ten- 
sion on the one will be 10 lbs. — but on the part back 
of the first, it will be the same as before — which 
was evidently less than 10 lbs. If the two sleds are 



26 KEY AND SUPPLEMENT 

of equal weight the tension on the connecting cord 
will be 5 lbs. I have heard students assert that 
no tension could be produced unless there were a re- 
sisting force — showing that they had not yet a correct 
conception of the relation between forces and masses. 
It requires force to move a free body. In such cases 
I have asked them to conceive that a cord were at- 
tached to the moon, and that they pulled upon it ; 
when they will severally admit that a pull of ten or 
more pounds may be easily exerted. If the sleds 
were not very heavy, it would not be possible for the 
boy to run sufficiently fast to maintain a constant pull 
of 10 pounds for a long distance; but if it can be 
done for a few feet only, it will answer the purpose 
of the illustration. 
7. ~No tension. 
Page 23, Art. 48. — The so-called Three Laws of Motion 
did not spring suddenly into philosophy. There was 
a long period of darkness succeeded by twilight and 
dawn before the truth shone out clearly. The prin- 
cijples of the three laws were known, and to a consid- 
erable extent realized, before Newton's time, but per- 
haps they had not been so clearly and sharply defined 
by any preceding writer, and much less had they been 
made the foundation of mechanical science ; hence 
there is a certain propriety in calling them Newton's 
Laws. Correct notions in regard to these principles 
date from the time of Galileo. Prior to his day, the 
leading philosophy was Aristotelian. Aristotle nour- 
ished between 300 and 400 years before Christ. In 
his philosophy there was no distinct idea of force as a 
cause, much less any idea of a relation between the 



TO ELEMENTARY MECHANICS. 27 

cause of motion and the momentum produced. He 
taught that a heavy body would fall faster than a 
lighter one — that when a body is thrown by the hand 
it ought to cease to move as soon as it left the hand 
were there no surrounding impulses, but that it con- 
tinued to move because the hand sets in motion the 
air about the body, and that the air acted afterwards 
in impelling the body. lie divided motions into 
Natural and Violent ; the former of which is illustra- 
ted by a falling body, in which the motion is constantly 
increasing, and the latter by a body moving on the 
ground, where the motion is constantly decreasing. 
It must not, however, be inferred that philosophers 
had no idea of cause and effect. Some general notions 
of this kind have always been entertained. 

Between the period of Aristotle and Galileo, many 
important principles were established. Archimedes 
(born 287 is. c.) developed some important properties 
of the Centre of Gravity, established the principle of 
the Lever, and some of the principles of Hydrostatics. 
History seems to show that the advanced position se- 
cured by this eminent philosopher was not main- 
tained, and that little or no advance was made until 
the time of Stevin — or Stevainus, as commonly writ- 
ten (1548-1620). His determination of the condi- 
tions of equilibrium of the inclined plane is so in- 
genious, it is worth repeating. Consider two inclined 
planes having a common vertex and horizontal base. 
Conceive a uniform long chain to be placed on them, 
and joined underneath so as to hang freely. He 
showed that it would hang at rest without friction, 
because any motion would only bring it into the same 



28 KEY AND SUPPLEMENT 

condition in which it was at first. The part hanging 
below would evidently be in equilibrium by itself, 
hence if that part be cut off the remaining part will 
be at rest ; hence the condition of equilibrium is — 
the weights on each part must he exactly proportional 
to the lengths of the planes. If one side becomes verti- 
cal the same proportion holds true. 

Galileo forms the grand connecting link between 
the philosophers of the ancient and modern physical 
sciences. He was born at Pisa, February 18th, 15G4, 
39 years before the death of Michael Angelo, and 21 
years after the death of Copernicus, and died on the 
8th of January, 1642, the year in which Sir Isaac 
Newton was born. The science of motion began with 
him. He taught that motion was due to force — that 
all bodies in a vacuum would fall with equal velocities 
— that inertia of matter implies persistence of condi- 
tion — he gave a satisfactory definition to momentum 
— also stated with approximate precision the princi- 
ple that " action and reaction are equal " — also estab- 
lished the principle of " virtual velocities,'' which 
was made by Lagrange to include all of mechanical 
science in one expression. He made a mathematical 
analysis of the strength of beams, of projectiles, of 
the pendulum, of floating bodies, and of the inclined 
plane. For his investigations in other fields of sci- 
ence — see some biographical sketch. 
Page 23, Art. 49.— First Law of Motion. Says Whew- 
ell, in his History of the Inductive Sciences : "It 
may be difficult to point out who first announced this 
Law in a general form." We have already seen that 
the facts involved in it were recognized by Galileo. 



TO ELEMENTARY MECHANICS. 29 

It is equivalent to saying that every change is due to 
a cause, and yet, to cover the entire ground, this state- 
ment needs modification. Motion is due to a cause, 
but when the cause ceases, the motion of a free body 
does not cease, it simply becomes uniform. Change 
of position is not then necessarily due to a coexisting 
cause, but may be due to a cause remote in time. 
Change of condition , however, requires a present act- 
ing cause ; and the latter will produce either a change 
in the rate of motion, or of the direction of motion or 
of both. 

The law cannot be proved by direct experiment, 
for it is practically impossible to remove from the 
body all acting forces, and hence uniform motion un- 
der the action of no forces is not realized by experi- 
ment. It may, however, be observed that the less the 
resistance the more nearly uniform will be the motion, 
and hence we are led to infer that if all resistance 
could be removed, the motion would be strictly uni- 
form. Similarly, it is observed that a body projected 
on a very smooth plane moves so nearly in a straight 
line that we are led to infer that if there were no de- 
flecting causes the path would be exactly straight. 
The law is the result of induction rather than of 
proof, but it appears go perfectly reasonable that we 
assent to it as soon as it is properly illustrated. The 
strongest proof of its correctness lies in the fact that 
deductions founded on this hypothesis agree with 
the results of observation. 

The process of induction consists in conceiving 
clearly the law, and in perceiving the subordination 
of facts to it. 



30 KEY AND SUPPLEMENT 

Page 23, Art. 50. — Second Law. Change of motion is 
hi proportion to the acting force. If all bodies were 
of equal size it would only be necessary to consider 
their relative velocities in determining the effect of 
forces. But a larger body requires more force than a 
smaller one of the same substance to produce the same 
velocity under the same circumstances ; in other words 
— mass and velocity are both involved — and the term 
motion here means momentum. This also agrees 
with Newton's explanation of the term. It is better, 
therefore, to word the law thus : Change of momen- 
tum is in proportion to the acting force. It must be 
particularly observed that the law does not assert that 
momentum varies as the force • but that it is a change 
of the momentum that varies as the force. 

This law was clearly perceived by Galileo, and by 
means of it and the first law, he determined that the 
path of a projectile in a vacuum was a parabola. The 
law, however, was not considered fully established 
until the theory in regard to the motion of the earth, 
involving both this law and the law of universal grav- 
itation, was realized. The triumph of both laws was 
complete at the same time. 

Page 24, Art. 51. — Third Law. Action and Reaction 
are equal and opposite. The first and second laws 
refer to one body only ; this involves two bodies. It 
asserts that an action between two bodies is of the 
same intensity upon each, but that the direction of 
action upon one is directly opposed to that upon the 
other. Every action implies an equal opposite action 
— one being called a reaction in reference to the 
other. No force, acting in one direction only, exists ; 



TO ELEMENTARY MECHANICS. 31 

it is always accompanied by an equal but opposite ac- 
tion. Eo force is known to exist without the pres- 
ence of matter. Force does not act in curved lines ; 
the action and reaction between two particles is in 
the right line adjoining them. 

Kewton gave three examples illustrating this law : 

If any one presses a stone with his finger, his fin- 
ger is also pressed by the stone. 

If a horse draws a stone fastened to a rope, the 
horse is drawn backward, so to speak, equally towards 
the stone. 

If one body impinges on another, changing the 
motion of the other body, its own motion experiences 
an equal change in the opposite direction. 

It does not seem rational that the stone will push 
the finger in the same sense that the finger presses 
the stone. The finger appears to be an active agent 
while the stone is inert. In the strictest sense we 
should say that, in the attempt to press a stone a 
force is developed between the finger and the stone, 
which force acts equally in opposite directions ; in 
one direction against the finger, in the other against 
the stone. 

Similarly, in regard to the horse and stone. In the 
former example the condition is statical, but in this 
the horse is supposed to move the stone. The horse, 
evidently, is not actually pulled backward, although 
there is an actual backward pull upon the horse by 
the rope. The fact is, that, in the effort to draw the 
stone, a force is developed w r hich produces tension in 
the rope, which tension acts to pull the stone one 
way, and the horse the opposite way. As the horse 



32 KEY AND SUPPLEMENT 

is able to take a footing on tlie earth, he is able to 
exert a force on the rope equal to," or exceeding, that 
necessary to overcome the friction of the stone, and 
thus move the stone. The pressure between the horse 
and the earth also acts in opposite directions. 

In the third illustration another idea is presented. 
Motion is used in the sense of momentum, as before 
stated, and it should read, the cltange of momentum 
in tiuo impinging bodies is the same in loth ladies 
but in opposite directions. This is a necessary result 
of the second and third laws. The forces being 
measured by the change in the momentum, and the 
pressures being equal between the impinging bodies, 
it follows that the changes in their momenta must be 
equal. Hence if one loses momentum the other must 
gain, and the loss in one case must equal the gain in 
the other. 

Thus if the body whose mass is 2[ x impinges upon 
another whose mass is M 2 , v t and v 2 , their respect- 
ive velocities before impact, and v\ and v' 2 their re- 
spective velocities at any instant after the first contact, 
and assuming, as we will in establishing the formula, 
that the velocities are in the same direction, in which 
case it will only be necessary to change the sign of one 
of the velocities if they move in opposite directions ; 
then will the momentum of M x before impact be 

and after impact 

and as 2T 1 is supposed to be the impinging body it 
will lose velocity, and we have for the momentum 
lost by M t 



TO ELEMENTARY MECHANICS. 33 

31{i\ — M x v\. 
Similarly, the momentum gained by M 2 will be 

3f 2 V r 2 ~ -2^8) 

which, according to the third law, must equal that lost 
by the former body ; hence for all stages of the mo- 
tion after the first contact, not only during compres- 
sion, but also at and after separation, we have 

Mfa - v\) = M 2 (y' % - <y 2 ). 

If M 2 be the impinging body, we have 

which easily reduces to the former equation. If they 
move in opposite directions before impact make i\ or 
v 2 negative, the impact is here supposed to be direct 
and central, for which case the equations are true 
whether the bodies be elastic or non-elastic. Several 
cases are discussed on pp. 85-90 of the text. 

(It appears that Whewell, in his History of the Inductive 
Sciences, has not drawn a sufficiently clear distinction between 
the second and third laws. He appears to hold that the third 
law gives a measure of the pressure or force, whereas the sec- 
ond law is the only one of the three that gives it). 

Page 24, Art. 52. — We are not informed who first 
gave the laws for the composition and resolution of 
forces ; but Galileo was one of the first to make use 
of them in explaining curvilinear motions. The 
method was systematized by Descartes by the aid of 
the systems of rectilinear axes. 



34 



KEY AND SUPPLEMENT 



exeecises. 
Page 26. 

1. Because the force of gravitation draws it from 

the rectilineal path in which it was projected. 

2. The least force. 

3. 500 pounds. 
Page 27. 

4. Midway between their initial positions. 

5. He must aim to walk southwesterly. To find the 

direction, draw a line, AB, of any 
length to represent the one mile 
due east, and a line, A C, perpendic- 
ular to AB, and of such length 
that the hypothenuse, BO, will be 
three times as long as AB ; then 
will the angle ^LB 6' represent the 
required direction. 
We have 



cos B = i ; 
3 ' 

.'. ABC = 40° 32° 
Make AB = 3, the A 
angle BBC = 45°, 
and BC equal to 
8; join AC, then 
will AC represent 
the resultant direc- 
tion. A numerical 
solution gives I) AC 
= 33° 10' and the 
course will be S. 56° 50' E. 





TO ELEMENTARY MECHANICS. 35 

BD= DC=8 sin 45° = 8 x I V2 = 4 V~2 

AD = 3 + 4 V2 

' nAn 4 a/2 32 - 12 V2 

tan. i>^l U = — = = 

3 + 4 \/2 23 

0-6535; .-. Z?^<7= 33° 10'. 

The velocity will be 10-34 miles per hour. 

7. It will be 10 x cos 45° = -J- V% x 10 = 5 x 

1.4142 + =7.0710 +. 

8. Yes, and will move towards the rear end of the 

car (First Law). Because it tends to preserve 
the velocity which it had just before the col- 
lision. 

9. In the first edition F % was 30 lbs. It was in- 

tended to be 20 lbs. so as to show that the re- 
sultant of two velocities may be the same as 
those producing it. In this case the triangle 
of velocities will be equilateral, and hence 
each of the angles will be 60°. 

If 20 and 30 pounds be used, we have for 
the diagonal of the parallelogram 



E = V900 + 400 + 2 x 20 x 30 x 0-5 
= 43-59 lbs.; 

and for the angles 36° 35' 12", and 23° 24' 48" 
respectively. 
Page 27, > Arts. 59, 60. — The intensity of a force is 
known only by its results. 

Page 29. 

Art. 63. — The case of a force acting normal to the 
path of a body was a difficult one with the philoso- 



36 KEY AND SUPPLEMENT 

pliers living about Galileo's time, and is not the 
easiest to explain by elementary processes at the 
present time. It will be considered in Chap, xvi., 
p. 220 of the text. 

Art. 64. — We say that gravity tends to draw bodies 
towards each other ; but we only know that it 
causes them to move towards each other when free. 
It is quite as proper to speak of its pushing as of 
drawing them towards each other. 

Some attempts have been made to explain the 
essential nature — or the cause — of gravitation. One 
of the most celebrated of these theories was given 
by one Le Sage. According to his theory, lines of 
force acted in nil conceivable directions through 
space, and as two bodies intercept those lines which 
would pass through both bodies, there would be 
more force to drive them towards than from each 
other. (See Theories of Gravitation by ¥m. B. 
Taylor, of the Smithsonian Institute, Washington, 
D. C.) But no theory thus far suggested is con- 
sidered sufficient to account for the fact of gravi- 
tation. 

Page 29, Art. 65. — The law of universal gravitation is 
one of the discoveries which aided in immortalizing 
the name of Sir Isaac Newton. He first conceived the 
nature of the law, and then proceeded to prove it. He 
assumed that if the law was correct it ought to explain 
the circular motion of the moon about the earth — in 
other words, that the pulling force (so to speak) of 
gravity at the moon would be just sufficient to draw 
it the required amount, from a tangent to the orbit 



TO ELEMENTARY MECHANICS. 37 

If g be the acceleration of a pulling body at the sur- 
. face of the earth, at the moon it will be g -i- D 2 — 
where D is the number of diameters of the earth be- 
tween the center of the earth and center of the moon, 
and is about 60-3612 ; and as g = 32-216 ft., we 
would have at the moon, the acceleration 32-246 -5- 
(60-3612) 2 = 0-0088 +. The force at the moon 
which would produce this acceleration equals the cen- 
trifugal force, and is given by the last equation of 
Art. 315 of the text, and is 

Force = m - w , 

where r is the radius of the orbit, T the time of a 
complete revolution, and m the mass of the body. 
But the force divided by the ?nass equals the accel- 
eration — as is shown by the last of the equations of 
Article 86 ; 

7 ,. Force 4:n 2 r 

.'. acceleration = = __ ■ , 

m 1 2 

which applied to the moon, and reduced, gives 0-0089 
+ (see Art. 319). The two results should agree if the 
law and data are both correct. It will be seen that 
the value of the radius of the earth enters into the 
computation, the correct value of which was not known 
to Newton at the time of his first investigations. 
Some fifteen years after he began the investigation, 
while attending a lecture in London, he obtained the 
correct value of the radius, with which he proved the 
truth of his proposition. 



38 KEY AND SUPPLEMENT 

The analysis by which the above result is reached 
is anticipated, but it will enable the reader to under- 
stand why an error in the true value of the radius of 
the earth vitiated the first result. 

It is questionable whether the story — that ISfewton 
conceived the law of gravitation by seeing an apple 
fall from a tree — so often taught to juveniles, is not 
purely fictitious. It is certain that he did not con- 
sider the law established for fifteen years after he 
first conceived it, during which time, it is said, he re- 
viewed his work many times. 
Page 30, Art. 67. — A history of pendulum experiments 
would furnish material for a book. The mathematical 
pendulum is an ideality, but a very useful one in dis- 
cussing the subject. Compound pendulums are 
necessarily used in making experiments. The most 
practical method of determining the acceleration due 
to gravity is by means of a pendulum ; and some of 
the results thus found are given on p. 244 of the text. 
The length of the seconds' pendulum has also been 
used for determining the standard of linear measure. 
Thus, the English law requires that the length of the 
yard shall be to the length of the simple pendulum 
vibrating seconds at the Tower of London reduced 
to the level of the sea as 36 to 39-13908. (See Art. 
328.) 

Page 31, Art. 68.— The formula 

g = 32-1726 - 0-08238 cos. 2Z, 

is given in The Mecarvique Celeste, Tome iii. v., § 42, 

[2,049]. 



TO ELEMENTARY MECHANICS. 39 

examples. 

Page 35. 

tf _ 

1. h = 2^ ••• v = Vsp = V2 x 32i x 100 = 

80-20 feet. 



2. t—-— ——■ = 9-3 seconds. 
<7 oH 



3. t = ^ = 3-109 seconds. 



7* = vj- igt 2 = 100 x 3-109 - 1 x 32| x (3-109) 2 = 
155-4 feet. 

Page 36. 

4. Acceleration = 32-J- ft. per sec. 

= 32| x 60 = 1,930 ft. per minute! 

5. Acceleration per second = 32-16666 feet, = 32- 

1G6QQ -f- 3-28 = 9-807 + metres. For 4 sec- 
onds, 9-807 x 4 = 39-228 metres. 

6. h^=ig# + v t, (Eq. (6) p. 34.); 
.-. 120 = \ x *??- £ 2 + 25*. 

Solving for £ we find 

£ = 2-0627 + sec. 
Also (Eq. (11) p. 13.) 

v =v Q -f gt 



40 KEF AND SUPPLEMENT 

= 25 + ^ x 2-0627 

D 

= 91-35 feet. 
7. 150 = i x =^p £ 2 + 25* ; 



.-. 25 



150 i/1800 x 193 4- 22500 

ffVQ "■ r 



193 ' r 193 a 

■150+608-19 



193 

= 2-37 seconds. 

8. For the falling body (Eq. (2) p. 34) 

and for the body projected (Eq. (8) p. 34) 

h — vt— l(/l 2 . 
Eliminating 7i gives 

v = gt; 

but 

hence eliminating ? gives 
v* = 2gh; 

.'.v = V%h-g 



9. The sound will be T ——- seconds in returning; 
1130 

hence the time of falling will be 



TO ELEMENTARY MECHANICS. 41 



4 


li 
1130" 


-t 










f/|l 5 (Eq.(4)p. 


.34); 






-:!*■ 


7* \ 2 27* 

ii3oy ~ <? ' 








(1130)»(; 


L130 + 16 T V/ 
= 231-6 + feet. 


-16 


x(1130) 2 ; 



10. Let x = the distance upward from the lower 
point to the point of meeting ; then will the 
point of meeting be a — x from the upper 
point. If t be the time of meeting, we have 
for the falling body 

a — x = vt + ^gfi, 

and for the body projected upward 

x= Vt-\gi\ 
Adding we have 

a= (F+ v) t; 

,.t: Cl 



which in the second equation gives 
a ( t-t a 



-v^r- v V- l9 rT-v. 



42 KEY AND SUPPLEMENT 

The distance below the highest point will be 



TTi( F+ *TTi)' 



EXERCISES. 

Page 36. — 1. He will. To draw it at a uniform veloc- 
ity he must overcome the friction only, but at 
an increasing velocity he not only overcomes 
the friction but also exerts additional force to 
overcome the inertia of the mass (see Second 
Law). 

2. Attraction is more. 

3. Because the air resists less. 

4. .j-ly- ounce. 
Page 37. 

5. 10 inches. 

6. More, for the force of gravity is less at the equa- 

tor than at the poles. 

7. It will neither gain nor lose in weight if weighed 

with the same beam scales. It will lose in 
weight if weighed with the same spring bal- 
ance, for the resistance of the spring will re- 
main constant, while the force of gravity will 
be less. 

8. 9-S07 + metres per second. 

Page 38, Art. 78. — In the last two lines of the page 
it is assumed that the attraction of a sphere upon an 
external particle varies inversely as the square of the 
distance from the center of the sphere. We here 
submit a proof of the truth of the statement. New- 




TO ELEMENTARY MECHANICS. 43 

ton, in his Principia, proved the proposition of the 
statement geometrically ; we now use the calculus. 

Let ABB be a spherical shell, center (7, radius a, 
P the posi- 
tion of an ex- / ____— — ^\B 
ternal parti- 
c 1 e. Con- j»*- 
ceive two 
consecutive 

radial lines" X> 

drawn from P, cutting the shell in the points A 
and B. 

Proof. Let ds be an element of length of the circle at A, 
PA = r, PC = c, CB = a, k = thickness of shell, 8 = den- 
sity. Conceive a line joining Pand C, = APC, and y = per- 
pendicular from A upon PC. 
Then 

2 TtyJcds = volume of shell generated by the 

revolution of A about PC, 
2 TtydJcds = mass thus generated . 
The attraction upon the particle will be 
2 7c8kyds 
r 2 ' 
which resolved along the axis PC gives 

2 nSkyds 
^— cos 0. 



Let p = C 


'E = the perpendicular from C upon PB, then 




p = c sin ; . \ dp = c cos Q d ; 


also 


7'- — 2rc cos Q + c 2 = a 2 ; 




dr re sin Q 




' dQ r — c cos Q ' 


and 


ds ar 


dQ ~ r — c cos G ' 


(for 


ds 2 = dr* + rW). 



44 KEY AND SUPPLEMENT 



Hence 

2 Ttdkyd* r 2 itdku cos ar«M 
^ — cos = 



r — c cos 
2 Ttdkyadp 



cr (r — c cos 0) 

2 7r£&<2 pdp 
i — X 



fV - j 



which gives the attraction for a circular element of the shell ; 
hence for the entire shell we have 

4 TtSka C a pdp 



r. 



Va 2 - p* 

4 TCdkar 



hence for a constant radius a, the attraction varies inversely as 
the square of the distance of the particle from the center of the 
shell, which was to be proved. 

If c = a, we have 
4 rrdk, 

hence the attraction of a spherical shell upon a particle in its 
surface is independent of the dimensions of the shell. 

To find the attraction of a homogeneous sphere upon an exter- 
nal particle, make k = da and integrate, and we have 



(ah 
o 



4 Ttd r« . . 4 Ttda* 

a-aa = = — 

3 c- 



8 
= volume X — , 



Page 39. 

Art. 79. — Weight, to an uneducated person, is con- 
ceived to be essential to matter. Such an one would 
doubtless assert that a body falls to the earth because 
it is heavy ; but to the student of mechanics, weight 



TO ELEMENTARY MECHANICS. 45 

is nothing but the measure of a force, the magnitude 
of which depends upon the quantity of matter con- 
stituting the body. 
Page 41. 

Art. 83. — The unit of mass might be the piece of 
platinum which is used as the standard pound (see 
Art. 33), but as we have occasion to compare the 
force of gravity at different places, and as the force 
of gravity at London is assumed to be 32-1- feet, we 
have chosen to consider the unit of mass as about 

-— - of the weight at that place. 
32| & * 

Page 42. 

Art. 85. — Density is sometimes used in the sense 
of specific gravity, and if the density of water be 
taken as unity, the specific gravity of any substance 
(compared with water) will equal its density. We 
prefer to make the definition conform to the sense 
used in mechanics. The following are the definitions 
given by several authors : 

By the density of a body is meant its mass or quantity of mat- 
ter compared with the mass or quantity of matter of an equal vol- 
ume of some standard body arbitrarily chosen. — Towne's Ele- 
mentary OTiem.y p. 29. 

Density is a term employed to denote the degree of proximity 
of the atoms of a body. Its measure is the ratio arising from 
dividing the number of atoms the body contains by the number 
contained in an equal volume of some standard substance whose 
density is assumed as unity. The standard substance usually 
taken is distilled water at the temperature of 38°. 75 F. — Bart- 
lett's Elements of Analytical Mechanics,^. 29. 

Enfin si Ton represente par D la masse, sous T unite de volume 
du corps que Ton considere, D sera ce qu'on nomme la Densite 



46 KEY AND SUPPLEMENT 

de ce corps. On prend communement pour unite de densite celle 
de l'eau destillee a cette derniere temperature (4° du therniome- 
tre centigrade). — Poissox, Traite de Mecaniqae, jmge 108. 

The quantity of matter in a body does not depend on the size 
of the body only, but also on the closeness with which the par- 
ticles are packed. This difference is defined as a difference of 
density. Thus there is more matter in a cubic inch of lead than 
in a cubic inch of oak, and this is expressed by saying that the 
density of lead is greater than the density of oak. — Magnus, Les- 
sons in Elementary Mechanics, p. 60. 

Experiment shows that the weight of a certain volume of one 
substance is not necessarily the same as the weight cf an equal 
voluma of another substance. Thus seven cubic inches of iron 
weigh about as much as five cubic inches of lead. 

We say then that lead is denser than iron, and we adopt the 
following definitions. When the weight of any portion of a body 
is proportional to the volume of that portion, the body is said 
to be of uniform density. And the densities of two bodies of 
uniform density are proportional to the weights of equal volumes 
of the bodies.— Todhunter, Mechanics for Beginners, p. 7. 

The density of a body is the mass comprised under a unit of 
volume. — Silliman, First Principles of Philosophy , p. 67. 

Density is the quantity of matter contained in a unit volume ; 
the absolute density or the closeness with which the particles are 
packed being uniform throughout that unit volume. This def- 
inition is directly applicable if a body is homogeneous ; but if it 
is heterogeneous, and the density varies from point to point, the 
density at any point is the quantity of matter contained in a 
unit volume throughout which the density is the same as that at 
the point. Density is usually measured by means of comparison 
with some substance the density of which is assumed to be the 
unit-density. — Price,. Infinitesimal Calculus, p. 164. 

The quantity of matter in a body, or as we now call it, the 
mass of a body, is proportional, according to Newton, to the vol- 
ume and the density conjointly. In reality the definition gives 
us the meaning of density rather than of mass, for it shows \xz 



TO ELEMENTARY MECHANICS. 47 

that if twice the original quantity of matter, air for example, be 
forced into a vessel of given capacity, the density will be doubled 
and so on. But it also shows us that, of matter of uniform den- 
sity, the mass or quantity is proportional to the volume or space 
it occupies. — Thomson and Tait, Treatise on Natural Philoso- 
phy, p. 162. 

Heaviness (Fr., densite, Ger., dichtigkeit) is the intensity with 
which matter fills space. The heavier a body is, the more mat- 
ter is contained in the space it occupies. The natural measure 
of heaviness is that quantity of matter (the mass) which fills the 
unity of volume; but since matter can only be measured by 
weight, the weight of a unit volume, e. g. of a cubic meter or 
of a cubic foot of another matter, must be employed as a meas- 
ure of its heaviness. 

The product of the volume and the heaviness is the weight. 

The heaviness of a body is uniform or variable according as 
equal portions of the volume have equal or different weights. — 
Weisbach, Mechanics of Engineering, p. 160. 

The density of a, body is the degree of closeness between its 
particles. The term depends upon the hypothesis that the ulti- 
mate particles of matter have weight, and therefore mass pro- 
portional to their bulk. It coincides with specific gravity. — 
Peof. Nichol, Cyclopaedia of the Physical Sciences, page 177. 

On sait que le poids d'un corps varie avec l'intensite de la 
pesanteurmais que sa masse ne varie pas. Sous l'influence de la 
menie pesanteur, par exemple en un meme lieu du globe, le 
poids est evidemment proportiounel a la masse et le rapport des 
poids de deux substances sous le meme volume sera precisement 
celui de leurs masses sousle meme volume ; de la, la synonymie, 
qui existe entre les mots poids specifique et densite, qui ex- 
prime ses rapports. — Wtjrtz, Dictionnaire de Chimie. Sous Den- 
site. 

Tbe density of a body is the ratio of its mass to its volume.— 
Smith, Elementary Treatise of Mechanics, p. 44. 



48 KEY AND SUPPLEMENT 

exercises. 

Page 43. 

1. The matter outside of one-half the radius would 

produce no effect ; and that within would at- 
tract as it* it were all at the centre. The sphere 
of one-half the radius will contain one-eighth 
the matter, and the inverse square of the dis- 
tance will be 4 ; hence the weight will be 
| x 4 x 10 = 5 lbs. We get this result more 
directly by saying, as in Art. 78, that the at- 
traction will be directly as the distance from 
the centre. 

2. 10 -T- 2 2 = 2J- lbs. 

3. Nothing. 

4. At a distance from the outside somewhat less 

than half the thickness of the shell. The ex- 
act distance cannot be found unless the thick- 
ness of the shell be given. (If JR be the out- 
side radius, r t the inside, and r the required 
distance from the centre ; then 

*g(/'-n 8 ) . frrCK 8 -*'!' ) .. - . 10 
P • —& •• 5 • 10 

or 

2B 2 (r s — rfl = >' 2 (7Z 3 -n 8 ); 

which is a cubic equation, and may be solved 
by Cardan's method.) 
Page 44. 

5. Yes for the first answer ; No for the second. 

6. On the opposite side. He could not stop at the 



TO ELEMENTARY MECHANICS. 49 

centre by a mere effort of the will. Some entertain 
the idea that the will of a person, causing a movement 
of parts of the body, could, in a measure, control the 
movement of the body as a whole ; but, as a fact, the 
matter of the body is subject to the action of forces, 
like any other matter. Xo part of the body can be 
moved except in accordance with the three laws of 
motion. If an arm is moved in one direction, some 
other part of the body will be moved in an opposite 
direction (the body being free). While passing across 
the hollow referred to in the Exercise, the person 
might throw his arms about, or kick, or perform the 
evolutions of swimming, or rowing ; but as there is 
supposed to be no matter, or no body, for him to act 
against, he could not change his rate nor direction of 
movement. In drawing his feet forward he would 
necessarily pull some part of the body backward. It 
is shown, by higher analysis, that the centre of gravity 
of a moving system is unaffected by the mutual ac- 
tions of internal forces — so, in this case, the motion 
of the centre of gravity of the person would be en- 
tirely unaffected by any contortions the person might 
make. 

7. At the centre of the sphere. At the same point. 
Uniform. 

Page 43. 

8. The ball. If the ball were very small compared 
with the person, the movement of the person might 
be neglected compared with that of the ball ; just as 
the motion of the sun is neglected compared with 
that of the planets. In the case of the planets, the 
pulling force is dependent upon their masses and dis- 

3 



50 KB* AND SUPPLEMENT 

tances, but in the case of the person this force is de- 
pendent upon his muscular exertion. 

9. He could not. In the effort to throw the ball 
away, a force is developed between the hand and ball, 
which acts equally between the ball and hand, but in 
opposite directions in accordance with the third law; 
and hence the ball would move one way and the per- 
son the opposite way; and both would move in 
straight lines in accordance with the first law, and 
their relative velocities will be inversely as their 
masses in accordance with the second law. 

If the person were placed at rest in any position in 
the hollow and unable to reach anything, he could not 
turn over, nor change ends ; that is, if his head were 
towards the north and his feet towards the south, he 
could not so change as to have his head towards the 
south and his feet towards the north. Should he at- 
tempt to so turn as to bring his head towards the 
south, he would cause his feet to approach his head, 
and they would meet about half way. He might suc- 
ceed in kicking his own head, or, if he were very 
flexible, the head and feet might pass each other ; 
but the body could not turn over so as to change ends. 
Neither could he roll over. If one end of the body 
turns one way, the other end necessarily turns the 
opposite way. 

If a cat be held with her back down two feet or 
so from the floor, and dropped, she will strike on 
her feet ; how does she do it ? According to the 
principles of mechanics, if there were nothing for 
her to act against it would be impossible for her 
to turn, and she would necessarily strike on her back. 



TO ELEMENTARY MECHANICS. 51 

While experimenting, I was surprised to find how 
near the floor the cat might be held, and often 
apparently perfectly un watchful when dropped, 
and yet alight on her feet. Now the movement of the 
body should be accounted for on mechanical principles. 
Instinct operates quicker than reason, and it appears 
to be certain that the cat, instinctively, initiates 
a rotation of her body at the instant she is dropped. 
While it is difficult to see how muscular action can 
be quick enough to produce this result, yet I see no 
other way of accounting for the rotation. Suspend 
the cat with a string at each foot, then suddenly cut 
the strings, and she will rotate herself. It is also an 
interesting fact that she will strike on her feet if let 
fall several feet, say six or eight feet. Now if she had 
the same initial rotation when falling six feet, as 
when falling two feet, why would she not turn too 
far in the former case and strike on her side ? To ex- 
plain this, we here state a principle not yet proved in 
this course. If M rotating tody self -contracts, it will 
rotate more rapidly, hut if it self-expands it will ro- 
tate more slowly. The cat has the ability, within a 
limited range, of expanding or contracting the trans- 
verse dimensions of her body, and to that extent of 
regulating the amount of her rotation. 

Consider still further the relations of the man and 
ball in the ninth exercise. He puts his hand in close 
contact with the ball, but without grasping it, and 
they move away from each other, until both strike 
the walls of the hollow. Then suppose that the man 
springs for the ball and seizes it, and then springs 
towards the centre of the sphere, but before reaching 



52 KEY AND SUPPLEMENT 

the opposite side throws the ball in anger. If the 
ball goes in a direction perpendicular to the line of 
his body (or, generally, in any line not passing 
through the centre of his body), the man will be 
thrown into a rotary motion as well as a motion of 
translation, and he will inevitably perform somer- 
saults while backing away to the opposite side of the 
hollow. 

Page 44. 

10. JOG ~ 32J- = 3 t Vj = 3.109. 

1 1 n u M (is * 200 x 6 

11. Density = -=- ■= — = 3-109. 

Volume 193 x 2 

1n ir 2-2 B> x 5 x (> nta 

12. Mass = ™ = -342. 

J. Jo 

13. Density = \^ * b J =-00581. 

14. If the resistance of the air be considered it 
would. In the second case, it would not stop, 
but would go from surface to surface with the 
regularity of a pendulum. (See text, p. 249.) 
In the third case the velocity would be greatest 
at the centre, if the hole be a vacuum ; but if 
it be filled with air, the greatest velocity 
would be passed before tile ball reached the 
centre. 

Art. 86. The value of F— Iff is sometimes called 
the absolute measure of force, but nothing is 
gained by the term, except that it distinguishes it 



TO ELEMENTARY MECHANICS. 53 

from the mere stress which it would produce if no 
motion resulted. Some English writers call the 
value of i^when thus expressed The Poundal, but 
this term has not come largely into use. 

Observing that acceleration is the rate of increase 
of velocity (Key, p. 6), it follows that this value of 
F is the same as that given by Newton's second law, 
that the force is proportional to the change of mo- 
mentum produced. But we are confident that the 
constant use of the term acceleration for rate of 
change of velocity possesses great advantage, since 
rate of change is liable to be considered the same 
as actual velocity. Indeed some text books assert 
that the momentum of a body is a measure of the 
force acting upon it — and call it the second law. 
Now a body may have momentum when no force 
is acting upon it. It is the rate of change of 'mo- 
mentum that measures the force producing the 
change ; in other and better words, the mass into 
the acceleration. 

Observe that the establishment of this ecpiation 
contains a very important principle. There is, 
strictly speaking, no relation between pounds and 
feet ; but the ratio of two weights may be the same 
as the ratio of two linear measures. A ratio is an 
abstract number, and often serves to connect con- 
crete quantities, forming an equation. Thus, in 
this case, 

-777. = a ratio, a mere number, = — . 
IF g 

The equation being established, it is operated upon 



54 KEY AND SUPPLEMENT 

algebraically. This use of ratio has many appli- 
cations in physics. 

answers to examples. 

Page 47. 

i ir 7 u :/ f 2 x 200 ° x 193 x i\ 

% 1286-66 = 35-87 feet. 

- 1 6 x 2 x 500 12 

SO-42 or 80 f V feet. 

q ir 7 ' u 1/ /2 x (25 - 10) x 193 x 100\ 

3. Ffeo* = |/ ( 6x ; 500 ) = 

^ 193 = 13-88 feet. 

. w 100 x 100 01 OK _ 

4. /fyrce = — — - — j^r = 31 -2o R>s. 

<32 x 10 

solutions of problems. 

Page 50. 

4. The tension equals the weight, P, plus the force 

W ' — P 
which will produce the acceleration. 7= ^ a is the 

acceleration when P is raised vertically. The mass 
multiplied by the acceleration is the moving force, or 
p \y _ p w _ p 

— • jp p g ; hence the tension is P + ir -v, P 

2 Tf'P 
= -rp= 73. Similarly, it equals IF minus the accel- 



TO ELEMENTAKY MECHANICS. 



. , 7r W-P^ 2WP 
erating force, or II ^ ^- H = 



W+ P ~ W+P 

5. The effective moving force is W — T, lience from 

Prob. 2, W- T= --/. 

9 
Substitute/^ ig, and W-T=iW; 

If ascending, f- W = — /, or T - W = I W; 

T — W + the force which will produce the ac- 

W 

celeration = W -i %g = f IF. 

exam p les. 
Page 50. 

2P — P 

2 - * = * ZP +JT 9 * = i x H A x 25 --= 134* ft. 

H* x 25 + 2 x 10 Kft KO ' - 

~ rk o? o — if 50 == 52-55 Its. 

J |- 4 x 25 — 2 x 10 

2 ° * = * TFT^ ^ or -P = 47-57 fts. 
4 * - ' P - W a* or a - 2s(P + TF > 



56 KEY AND SUPPLEMENT 

= 2x6-8x9fr =64 . 6ft 

i x 4 

2xl0xP. 
o. 1-— p-- - 10 -, -^-T.1D- 

5 = ^10 4 P ; . p = 3 , lu 



10 = f-^i_lil ; .-. p - 10 11 )S . 





P 


+ 


10 


2 


X 


10 


x P 




i J 


+ 


10 


2 


X 


10 


x 7 J 



»=-prnr' •■■*=»■ 



If the tension exceeds 20 lbs. P will be nega- 
tive. 






,/ 2 x 1U x 22 

' ^ 9. y JJJ. 



= 2-G15 seconds 



2 x 

AXSWEIiS TO EXERCISES. 

Page 51. 

1. The balance will indicate no tension. This 
question was given, because the author sometimes 
found that, in a ease like Fig. 21, some students would 
assert that the tension of the string ought to be P ; 
but by taking an extreme case, like the one in the 
exercise, the fallacy would be apparent. One of the 
best conditions of mind for searching after truth is to 
be convinced of one's error. When one admits that 



TO ELEMENTARY MECHANICS. 57 

his position is erroneous, he is generally in a condi- 
tion to admit the truth. 

2. If the acceleration is increasing, the tension will 
exceed the weight ; if uniform it will equal the 
weight ; if decreasing it will be- less than the weight. 

3. It will be less than his weight while descend- 
ing, and greater if ascending. In both cases, if the 
motion be uniform, it will equal his weight. 

4. The tension will be less than his weight. 
Page 52. 

Art. 91. — It has already been stated that, in the 
relations between force and motion, we have four 
fundamental elements, force, space, time, and mass. 
Force and matter are so intimately connected that it 
is impossible to completely divorce them ; but force 
may be abstracted from space and time — not in the 
sense that it can exist without them, but in the sense 
that it may be considered independently of them, and 
when so considered it is called stress. But a force 
acting upon a body may move it through space, and 
the space may be considered independently of time. 
The product of force and space both considered in the 
same line, and abstracted from time, is work. 

The term originated from the grosser ideas of labor, 
but the definition given in the text is applicable to 
the most refined actions in mechanics. All known 
forces in nature are constantly working. . Thus, riv- 
ers wear the beds of their streams ; wind drives the 
sail, uproots trees, produces drifts of sand, etc. ; heat 
expands bodies, and may overcome -their cohesion, 
etc. 

According to the definition, a man merely snp- 
3* 



58 KEY AND SUPPLEMENT 

porting a weight cannot be said to work, and yet lie 
soon becomes conscious of fatigue. But a more criti- 
cal examination of his case shows that the weight is 
not strictly at rest. The natural action of his organ- 
ism, especially the beating of his heart, causes slight 
elevations and depressions of his load, so that he is, 
in the strictest sense, constantly laboring. 

The following are some examples, of the average 
work accomplished by a man under various condi- 
tions. 

WORK OF MAN AGAINST KNOWN RESISTANCES. 

hrs per ft. lbs. 
lbs. day. per day. 

1. Raising his own weight up stairs or 

ladder 145 8 2,088,000 

2. Hauling up weights with rope, and 

lowering the rope unloaded 40 6 648, 00J 

3. Li fting weights by hand 44 6 522,720 

4. Carrying weights up stairs returning 

unloaded 143 6 399.600 

5. Shoveling up earth to a height of 

5 ft. 3 in... 6 10 230,800 

6. Wheeling earth in barrow up slope 

of 1 in 12, one-half horiz. veloc. 0.9 

ft. per sec. , and returning unloaded. 132 10 356,400 

7. Pushing or pulling horizontally (cap- 

stan or oar) 26.5 8 1,526,400 

8. Turning a crank or winch IS 8 1,296,000 

9. Working pump 13.2 10 1,183,000 

10. Hammering 15 8? 480,000 

PERFORMANCE OF A MAN TRANSPORTING LOADS HORIZONTALLY. 

In- pet lbs. con'd 
lbs. day. 1 ft. 

11. Walking unloaded, transport of own 

weight 140 10 25,200,000 



TO ELEMENTARY MECHANICS. 59 

hrs. per lbs. con'd 
lbs. day. 1 ft. 

12. Wheeling load in 2- wheel barrows,re- 

turn unloaded 224 10 13,428,000 

13. Wheeling load in 1-wheel barrow, re- 

turn unloaded 132 10 7,920,000 

14. Traveling with burden 90 7 5,670,000 

15. Carrying burden, returning unloaded. 140 6 5,032,800 
See Rankin e's Steam Engine, pp. 8-4-85, where the 

rate of doing the above works is also given. 

Morin and "Weisbach give 2,387 ft. lbs. per minute 
as the work which a man is capable of doing when 
working eight hours consecutively. This equals -££$fc 
= 0.07 4- HP. nearly. 

At an experiment made at Dresden in 1 880, men 
working only 2 minutes at a time on a hand fire-en- 
gine did 0-277 H.P. — or nearly four times that given 
above. 
Page 54, Art. 98. — Kystrom asserts in his writings, 
that work is not independent of time, for it requires 
time to move a body over space ; also that if one 
horse drew twice the load over the same space as an- 
other, he did twice the work and was twice as effi- 
cient. But Ave hold, and trust we have clearly shown, 
that time is properly abstracted in the idea of work- 
that efficiency is very different from work. We have 
also shown that if time is considered even implicitly, 
the velocity must also be considered. 

Art. 99. — The simple definition— Power is rate of 
doing work — is coming more generally into use. 
Page 56, Art. 104. — Friction is a force ; its value can 
be measured in pounds. It does not directly produce 
a positive acceleration, but a negative one. Its direct 
office is to destroy motion ; not to produce it. But 



60 KEY AND SUPPLEMENT 

indirectly it may produce motion by produc ng Lent. 
Heat produces motion, and the work of friction has 
its equivalent in heat, and this heat if collected would 
produce the same motion as that which it has de- 
stroyed. But in practice it is so quickly dissipated 
that, in most cases, it is apparently lost. 

A smooth surface is one from which the idea of 
roughness is abstracted. 

Page 58, Art. 107. — The laws of Morin are only ap- 
proximately correct. In machinery, the character of 
the surfaces in contact, the mechanical execution of 
the fitting up, and of the lubricants, are each and all 
important elements. See practical treatises and arti- 
cles upon the subject. Pankine, Steam Engine ', pp. 
14-18, Thurston, Friction and Lubrication. 

Page 59, Art. 109. — The frictional resistance of rail- 
road trains is principally rolling* friction under good 
working conditions. A railroad train in good order, 
and on a good road, will not be safe against starting 
under the action of gravity alone, unless the gradient 
is less than eighteen or twenty feet to the mile ; once 
started it will continue in motion on gradients as low 
as thirteen feet to the mile. The coefficient of rolling 
friction for trains in good order is ¥ |4o ^ 0-0025, or 
less than six pounds per ton. The resistance at start- 
1#n g is T || F = 0-0038 or %\ pounds per ton. 

The resistance of a locomotive is about 12 pounds 
per ton. 

The resistances on railroads, under average condi- 
tions, and including all forms of resistances, are given 
by Clarke. 



TO ELEMENTARY MECHANICS. 61 

AY lien the permanent way is straight, rails dry and 
clean, he gives for trains only 

for engine and train 

*=* + $?> 

where R is in pounds per ton gross, and v the velocity 
of the train in miles per hour. (JSLanual of Rules, etc., 
p. 965.) A Mr. Hughes found on an English "tram- 
way " a resistance of twenty-six pounds per ton. 

On railroads, frictional resistances are sometimes 
greatly increased by the resistance of the air, called 
"head resistance," and amounts, in pounds per square 
foot of front exposed, to 0-005 of the square of the 
velocity in miles per hour with which the air meets the 
head of the train. Side winds often increase the 
flange resistance seriously. 

The value of the coefficient of friction on ordinary 
railroads is 0-003, on well laid railroad tracks 0-002, 
on best possible railroad 0-001. 

Mr. S. Whinery (Trans. Am. Soc. C. E., April, 
1878) gives a formula for the total resistances of a 
train running on curves, 

where R = total resistance., D = degrees of curvature, 
g = gauge of track, t = length of rigid wheel base, a 



62 KEY AND SUPPLEMENT 

and n are quantities expressing resistances due to ac- 
cidental and irregular conditions. These resistances are 
inversely as the radius of curvature, directly as the 
load, and nearly independent of the velocity. 

Mr. O. Chanute (Trans. Am. Soc. C. E., April, 
1878) analyzes this increase of resistance as follows : 



Due to twist of wheel 0-001 

" slip " 0-1713 

" flange friction 0-2450 

" loss of couplings 0-0213 

Total 0-4386 



Loose wheels reduce this loss 20 or 25 per cent. The 
rigid form of wheel-base of European cars and loco- 
motives doubles the increase due to curves as well as 
increases the resistance 'on the straight line. Accord- 
ing to Mr. Chanute the " coning " of wheels increases 
the resistance from 0-125 to 0-25 pounds per degree 
of curve per ton. (Thurston, Friction and Lubrica- 
tion, pp. 13-18.) 

Recent experiments on the New York and Erie 
H. R. show that on a track of steel rails in first-class 
condition, the friction of a train at low velocities may 
be reduced to 3| or 4 pounds per ton on a horizontal 
road ; and that the rolling resistance on such a track 
in the summer may be safely taken at 5 pounds per 
ton. (R. E. Gazette, March 24, 1882, or HaswelPs 
Pocket- Book for Engineers, edition of 1882.) 



TO ELEMENTARY MECHANICS. 63 

Page 63. 

Example 3. — If /.i — 0, the expression /a?jW + h "IF be- 
comes It TT, in which case the work equals that neces- 
sary to raise the weight through the height of the 
plane. 



SOLUTIONS OF EXAMPLES. 

Page 64. 

1. It will raise 50 X 33,000 x CO ft. lbs. in 1 hour, 

which divided by 500 x 62-5 will give the 
cubic ft. = 3,168. 

2. The average height to which the material is 

raised will be 10 ft. Hence the work = 140 
(i n x 3 2 x 20) x 10 =r 197,920-8. lbs. 

Q F + F' 1000 + 200 1Q _ . 

,. V = ~2F * =7 2000 X 12 = 7 ' 2 m ' 

4. 39-37 inches = 3-2808 ft. x 2-2 lbs. = 7-217 

+ ft. lbs. 

5. 43,333 x 7-217 + *■ (32;808) 2 = 29,057 ft. lbs. 

6. Substituting in the answer to Prob. 2, p. 61, we 

3 

have H. P. = 0-9114 x 1 x 12* = 37-89. 

7. Work of the fall = 2,000 x 8 = 16,000 ft, lbs. 

Let x = the distance driven, then 10,000# = 
16.000; .\x=l-6 ft. 

8. Find the velocity in feet per minute. We have 



64: KEY AND SUPPLEMENT 

2 x 5280 



00 



The horse-power == Fv -s- 



QQ AAA 200 x 2 x 5280 , . 
83 ' 000 ' 01 ' 00x33000 = 1 *- 

_ yftP - 2 (P + TF).9 _ if 1 x 9x8-2(8 + 40)4 
y - M - ,.# w ~ ~ i|3 x 9 x 40 




_ 1 P - fiW ft _ x 5 -2 5x0-15 
' ^ " r ' "PTTr ** " * X 5+25 

x lp x 25 = 1206-25 + 72 = 10-75 ft. 





ANSWERS TO EXERCISES. 

1. One pound raised one foot. Work is a com- 

pound quantity, compounded of stress and 
space. 

2. See preceding remarks, text, p. 56. 

3. It is dependent upon time only implicitly, and 

in the sense that motion requires time. But, 
strictly, time should be abstracted. 

4. Tang. 15° = 0-208. 

5. 5 lbs. 

G. Mechanical power is not work, but rate of doing 
work — just as velocity is not space, but rate at 
which space may be passed. Mechanical power 
involves a unit of time, but work dues not. 
Page 66. 

Art. 41. — The doctrine of energy is the grand, gen- 
eral principle of modern physics. All the changing 



TO ELEMENTARY MECHANICS. 65 

phenomena of nature are but manifestations of the 
transmutation o£. energy. Its principles are not de- 
duced by any system of mathematics, but by a long 
series of inductions. We accept its general principles 
without attempting a general demonstration. 

Even work in a higher sense is but a means of 
transmitting energy. Thus, a horse works by draw- 
ing a load, but it is simply a means of transmitting 
the energy possessed by the horse, first into energy 
stored in the mass of the load, and second into heat 
by means of the friction overcome. Still, work is not 
only a convenient, but a useful term. Work clone is 
one idea, energy produced as its equivalent is another 
— force and space are the elements of the former, 
mass and velocity of the latter. 

DEFINITIONS OF WORK AND ENERGY. 

" Work is the overcoming of resistance continually 
recurring along some path." — Bartlett's Elements 
of Analytical Mechanics, p. 26. 

" Work consists in moving against resistance. The 
work is said to be performed, and the resistance 
overcome." — Rankin e's Applied Mecha?iics, p. 477. 

" Work is the effect of strain and motion combined. 
— Traetwine's Engineers 1 Pocket-Book, p. 445. 

Remarks on above from p. 446 same book : " Grav- 
ity acting in a body falling freely in a vacuum, and 
consequently unresisted, exerts no effort upon it, it 
neither goes before, and pulls it along, nor behind, 
and pushes it ; for there can be no pull or push except 



QQ KEY AND SUPPLEMENT 

when there is some force to pull or push against. 
But it simply, as it were, animates the body, or en- 
dows it with the power of locomotion. As the body 
falls, the force of gravity which gives it motion all 
remains unimpaired, and stored up in it, ready to ex- 
ert an effort against any other force which it may 
chance to meet with. Therefore a body falling unre- 
sistedly has no weight ; for gravity, which gives it 
weight alone while at rest, now gives it motion 
alone." 

(The word strain in the above definition is improp- 
erly used for stress. — Author.) 

" A force is said to do work if its place of applica- 
tion has a positive component motion in its direction; 
and the work done by it is measured by the product 
of its amount into this component motion." — Thomson 
and Tait, Nat Philos., p. 176. 

" Work done on a body by a force is always shown 
by a corresponding increase of vis viva, or kinetic en- 
ergy, if no other forces act on the body which can do 
work or have work done against them. If work be 
done against any force, the increase of kinetic energy 
is less than in the former case by the amount of work 
so done. In virtue of this, however, the body pos- 
sesses an equivalent in the form of potential energy, 
if its physical conditions are such that these forces 
will act equally, and in the same directions, if the 
motion of the system is reversed." — lb., p. 177. 

" An agent is said to do work when it causes the 
point of application of the force it exerts to move 
through a certain space. Motion is essential to 



TO ELEMENTAR5T MECHANICS. 67 

work." — Twisden, Elementary Introduction to Prac- 
tical MecJianics, p. 18. 

" Mechanical effect, or work done, is that effect 
which a force accomplishes in overcoming a resist- 
ance. It depends not only on the force, but also on 
the space during which it is in action, or during 
which it overcomes a resistance." — Weisbach's Me- 
chanics of Engineering, p. 168. 

" Work is the production of motion against resist- 
ance." — Todhuntek, Mechanics for Beginners^. 337. 

"Work, same as before. According to this defini- 
tion, a man who merely supports a load does not 
work; for here there is resistance without motion. 
Also while a free body moves uniformly no work is 
. performed ; for here there is motion without resist- 
ance." — Todhuntek, Nat. Philos. for Beginners. 
p. 255. 

" Whenever a body moves through any space in a 
direction opposite to that in which a force is acting 
on it, work is said to be performed. It is evident 
that the application of force is necessary to overcome 
resistance, and it is very often found convenient to 
measure the work done by the amount of force ex- 
pended, and the distance in the direction of the force 
through which it has been employed."— Magnus's 
Elementary Mechanics, p. 102. 

" Thus the increase of vis viva, which is also the 
work done by the acting forces on the body." — 
Price's Infinitesimal Calculus, vol. iii., p. 636. 

" Work is done when resistance is overcome, and 



g KEY AND SUPPLEMENT 

the quantity of work done is measured by the prod- 
uct of the resisting force and the distance through 
which that force is overcome." — Maxwell's Theory 
of Heat, p. 87. 

" Work is the overcoming of mechanical resistance 
of any kind." — Nystrom's Pamphlet on Force of 
Falling Bodies, etc., p. 25. 

DEFINITIONS OF ENERGY AND YIS YIYA. 

" Energy expresses power to do work, or force stored 
and ready for use." — McCulloch's Treatise on Me- 
chanical Theory of Heat, p. 40. 

" Vis viva (energy) is a quantity which varies as 
the product of Jthe mass of a particle and the square 
of its velocity."— Prices Infinitesimal Calculus, vol. 3, 
p. 380.— 2d Ed. Oxford. 

" Living force, or vis viva (or energy), is nothing 
more than an expression referring to the quantity of 
work (motion and strain combined) which the force 
in a body at any given instant could perform, if left 
to itself, without afterwards receiving any additional 
force."— Trautwine, Civil Eng. Pocket Book, p. 446, 
Ed. 1872. 

" Energy measures the quantity of working power 
of a moving body." — Bartlett, Elements of Aiuilyt. 
Mech., 9fch Ed*, p. 116. 

" The product of the mass of a body by the square 
of its velocity is called its living force or vis viva" — 
Mecanique Celeste, p. 99. 



TO ELEMENTARY MECHANICS. 69 

"Energy means capacity for performing work." — 
Rankixe, Applied Mechanics, p. 477. 

" Vis viva-, or living force (or energy), is the power 
of a moving body to overcome resistance, or the 
measure of work which can be performed before the 
body is brought to a state of rest." — Silliman, Prin- 
ciples of Physics, 2d Ed., p. 78. 

"Energy is the capacity a body has, when in a 
given condition, for performing a certain measurable 
quantity of work." — Todhunter, Natural Philosophy 
for Beginners, p. 264. 

"Energy, of a body is power of doing work." — 
Magnus, Lessons in Elementary Mechanics, p. 110. 

" Energy of a body is the capacity which it has of 
doing work, and is measured by the quantity of work 
which it can do. The kinetic energy of a body is the 
energy which it has in virtue of being in motion." — 
Cumming, Theory of Electricity, p. 5. 

" Kinetic energy or vis viva is denned as half the 
product of the mass into the square of the velocity." — 
Ibd., p. 13. 

" Energy is defined to be capacity for doing work." 
" It is of two kinds — kinetic or actual when the 
body is in actual motion. Potential or latent when 
the body, in virtue of work clone upon it occupies a 
position of advantage, so that the work can be at any 
time recovered by the return of the body to its old 
position." — Ibd., p. 14. 

"Energy is the capacity of a body to perform 



70 KEY AND SUPPLEMENT 

work. Energy is said to be stored when this capacity 
is increased, and to be restored when it is diminished. 
The units of work and of energy are the same." — 
Weisbach, Meclt. and Fng. Translator's note, bottom 
of page 168. 

" If we adopt the same units of mass and velocity 
as before there is particular advantage in denning 
kinetic energy as half the product of the mass into 
the square of the velocity." — Routh's Rigid Dynam- 
ics. 

Page 07. 

Akt. 112. — If the body had an initial velocity v , 
the work done upon it in passing from a velocity 
v to -y, or the work which would be given out by it 
in passing from the velocity v to v , will be 

lM?-lM K * = iM(* -»„*)■ 

This may be deduced in another way, if we antici- 
pate the equations for momentum. 
Thus, Eq. (2), p. 78, of the text is 

Ft - M{v - v Q ). 

The space over which F acts will be equivalent to 
the average velocity into the time, or 

s = l(v+v )t. 

Multiplying these equations, member by member, 
and canceling t, gives 

Fs = \M(i? - v Q % 



TO ELEMENTARY MECHANICS. 71 

as before. But we do not consider this method as 
good as the h'rst, for when the velocity changes irreg- 
ularly, it is not so evident that the space equals the 
average of the extreme velocities into the time. 

Art. 113. — Potential energy is relative. Thus if a 
body whose weight is 10 lbs. is 40 feet above the 
earth, its potential energy in reference -to the earth 
is 10 x 40 = 400 ft. lbs. ; but if it be over a well 20 
feet deep, the potential energy in reference to the 
bottom of the well, will be 10 x 60 = 600 ft, lbs. ; 
and in reference to its own position it is nothing. It 
is the work which the body may do in reference to 
some point — or condition — arbitrarily chosen. 

Page 69. 

Art. 115. — For the mathematical theory of heat, 
see Poisson Traite des Gheleur, Fourier Theorie de 
la Cheleur, Rankine on The /Steam Engine, Clausius 
on The Theory of Heat, Hirn's investigations, Max- 
well's Theory of Heat, McCulloch's Theory of Heat, 
Tait's History of Thermodynamics, etc., etc. 

Page 72. 

Art. 117. — This is one of the most important phys- 
ical constants which has been determined in recent 
times. For a comprehensive and able review of the 
methods by which it has been determined, as 
well as for a description of that author's methods 
and results, see Mechanical Equivalent of Heat, by 
Professor H. A. Rowland. (Proceedings of the 
American Academy of Arts and Sciences, 1879, p. 
45.) Prof. Rowland's results differ by only a small 
per cent, from those given by Joule. He states, 



72 KEY AND SUPPLEMENT 

p. 44, that the value found by Joule at 14°, agrees 
with his results at 18° 0. The value 772 foot pounds 
still stands as a practically correct one. 
Page 73. 

Art. 118. — In regard to energy generally, it appears 
that all of it — or at least very nearly all — originates 
with the -sun. It was a beautiful remark of John 
Stephenson — as he saw a railroad train winding its 
way through the country — " that train is drawn by 
the heat of the sun." The heat and light of the sun 
caused the growth of vegetation ; that vegetation in 
time was gathered into great masses, which in time 
became coal iii the mine. This coal was brought 
forth and used as fuel in the locomotive; so that it 
originated in the action of the sun. All energy on 
the earth is due to the light and heat of the sun. 
Activity in the commercial world is directly depend- 
ent upon it ; for if the sun, on account of spots upon 
it, or from other causes, does not dispense with its 
usual heat, short crops will result, and thus affect all 
the business of a country, if not of the world. In this 
way the sun may be charged with causing — more or 
less directly — depressions in trade, or activity in com- 
merce, as the case may be ; and hence, in some cases 
at least, of producing sadness or cheerfulness in the 
home circle. Even the religious world is affected by 
the action of the sun. In earlier times, on account 
of the superstitions of the people, religious leaders 
appealed more or less effectually to the fears of their 
followers when the sun's rays were cut off by an 
eclipse; and in modern times the feeling of depend- 
ence upon the Creator is too often modified by the 



TO ELEMENTARY MECHANICS. 73 

prosperity or depression caused by the circumstances 
surrounding them, the conditions of which were de- 
pendent upon the elements of nature, and these 
caused, more or less directly, by the action of the sun. 
The sun appears to be dispensing its energy to his 
family of planets, and in this way wasting itself 
away. Sir Isaac Newton realized this condition of 
things and saw — or thought he saw — the necessity 
of the sun's being replenished in order to maintain its 
stock of energy ; and he conceived that this might be 
done by one comet after another falling into the sun. 
As the comets come from remote regions of space, 
they would possess a large amount of energy when 
they struck the sun, and by falling into it would pro- 
duce intense heat. No comet, however, has within 
historical times been known to fall into the sun ; but, 
on the other hand, the most critical examination of 
their orbits shows that their paths are nearly as well 
defined — and nearly as fixed in position — as any of 
the planets ; and no cause is known to exist that will 
cause them to fall into the sun. It was formerly sup- 
posed that the ether of space caused a resistance to 
the motion of all bodies in space ; and if it did, not 
only would the comets, but also the planets, ultimately 
fall into the sun. However, nothing — absolutely noth- 
ing — is Tcnown in regard to the effect of this ether - 
upon the motion of bodies in it. Comets and planets 
have moved in their orbits for untold ages, and, for 
aught we know, have maintained their relative posi- 
tions. If the comets were destined to fall into the 
sun, they must have been doing so for ages and ages, 
and hence must originally have been comparatively 



74: KEY AND SUPPLEMENT 

very numerous. It would seem that sufficient time 
had elapsed since the existence of the solar system 
to have exhausted this stock of energy — still, at the 
present time, comets are numerous. Similarly, in re- 
gard to the planets, the most refined observations, 
combined with the most refined analysis, have failed to 
detect any modification of motion due to the ether 
of space. 

More recently, the late Professor Benj. Pierce, of 
Cambridge, Mass., put forth the theory that the en- 
ergy of the sun was supplied by meteorolites falling 
from an immense distance directly into the sun. The 
meteors are dark bodies, and it is assumed that there 
may be multitudes of them scattered through space. 
That there are many, is shown by the fact that they 
frequently fall upon the earth. Admitting the truth 
of this hypothesis, it seems inevitable that, in the 
course of time, the supply will be exhausted — and then 
the question, What will follow? becomes a serious 
one to science. 

Space, which, in our younger days, we conceived 
to be void, is really filled with something, and there 
may be vastly more inert matter scattered through 
it than we have imagined. 

A mere contraction of the volume of the sun, 
caused by the mutual attraction of its own particles, 
will produce heat. The author has shown that if the 
earth contracted from twice its present diameter to its 
present size, and all the energy thus produced be 
changed into heat, and uniformly disseminated 
throughout the mass, the temperature would be raised 
44,655 degrees F., if it had the specific heat of water, 



TO ELEMENTARY MECHANICS. 75 

or 357,240 degrees F., if it had the specific heat of 
iron. (See Analytical Mechanics, p. 229, or Mathe- 
matical Visitor, 1880, p. 134.) 

We believe that the solar system is stable, that it is 
not made to run down, that it has the elements of 
self-preservation ; but we cannot prove it. Neither 
can those who entertain the opposite view prove their 
position. This problem is, at present, beyond the 
reach of science. 

solutions of examples. 

Page 75. 

25 

1. — - = the mass, and as 32J is the acceleration 
d2g- 

in feet per second, the velocity should be in 

the same units ; hence v — \%° = -f feet per 

second, and the work will be 

2. The work will vary as the depth, and the energy 

as the square of the velocity ; hence, 

depth ■=■ 2(V) 2 = 18feet. 

Page 76. 

3. Reducing the tons to pounds, we have 

60 tons = 60 x 2000 lbs. = 120,000 lbs. 

Similarly, 

, A ., , 40 x 5280 - x , 

40 miles per hour = — — - it. per second 

60 x 60 



76 KEY AND SUPPLEMENT 

The friction = 60 x 8 = 4S0 lbs. 
Let x = the required distance in miles = 5280a; 
feet, then the work will be 4S0 x 5280 x ; 
hence we have 

120000 /40 x 5280\ 2 . 
480 x 5280* = i • - 3 ^ • ^ 6 x 60" ; ' 

which solved gives 

, v __i0^_ (™\ l = 2-M miles. 

4 Let aj = the required number of pounds of wa- 
ter ; then the energy put into water in raising 
it from 32° F. to 212° will be 

772 (212 - 32) # = 772 x 180.r. 

The kinetic energy stored in the train will be 

200000/20 x 5280V. 
* ' ~32f \~60 x 60/ ' 

which, by the conditions of the problem, equals 
the heat energy to be put into the water; 
hence 

200000 /88\ 2 . 
772 x 180^ = \ x -321" \%) ' 

... «? = 19-2 lbs. 

Page 76. 

p _ , iW 4 - 0-2 x 10 193 __ 193 

5 '/-pTT^ = IT" - x -6"-"l2 



TO ELEMENTARY MECHANICS. 77 

also for the velocity 

v> = 2/s = 2 x 4-6 x 5 = 46. 
The work will be 

10 x /* x distance, 
and the energy will be \Mv % ; 

.-. 0-2 x 10 x distance = J x x 46 ; 

.*. distance = 3^- feet. 

6. The friction = 200 x 0-2 = 40 lbs. Let the 
velocity per minute be a?, then the work per 
minute will be 40a?, and for 3 minutes it will 
be 120a?. The energy of one pound of water 
raised one degree F. is 772 ft. lbs., and of 5 lbs. 
it will be 5 x 772, and for 50 degrees it will 
50 x 5 x 772. 
Hence 

120® = 50 x 5 x 772 ; 

.*. a? = 1,608 ft. per minute. 

• answers to exercises. 

Page 76. 

1. It is not; force is only one of the elements pro- 
ducing energy. 

2. Ability to do work. Work has, however, been 
done. 

3. It produces action of the stomach, thus in- 



78 KEY AND SUPPLEMENT 

volving energy ; promotes action of the heart ; causes 
the growth of the bone, muscle, and flesh, and these 
enable the animal to move about — to do work — to 
swallow more food — to lie down — to get up, etc., etc. 

4. Because, in the first place it is not as concentra- 
ted, and, in the second place, the heat is more quickly 
conducted away. 

5. It will. It is due to this cause that meteors are 
visible. The meteorolites which fall upon the earth 
have the appearance of having been partly melted, 
and hence must have been subjected to great heat. 
This is due to the compression of the air in front of 
the meteor and of the friction of the air against its 
sides as it passes swiftly through the air, and the heat 
thus produced is so great that the meteor is heated to 
redness, and thus appears like a shooting star, as it 
really is. It is probable that the smaller meteorolites 
become so nearly consumed by the great heat that 
they could scarcely be found after they had fallen 
upon the earth, but larger ones have been found 
which struck the earth with such violence that they 
nearly or quite buried themselves. It has been sug- 
gested that the great iron deposit in the upper pen- 
insula of Michigan, a short distance west of Mar- 
quette, was probably a large meteor, and a similar 
suggestion has been made in regard to the iron mount- 
ain in the State of Missouri. 

Page 77. 

6. The friction of the water produced by moving 
against the banks and bed of the stream produces 
heat, which, escaping into the surrounding air, modi- 
fies its temperature. 



TO ELEMENTARY MECHANICS. 79 

7. First by the heat clue to the friction, and second, 
by inducing a quicker circulation of the blood more 
heat is supplied to the parts. 

Page 77. 

8. 8--50. This exercise is intended to draw atten- 
tion to the fact that the value of wood for fuel de- 
pends upon its capacity for producing heat ; or, in 
other words, of its inherent heat energy. If the heat- 
ing power of a given weight of hickory be 100, it has 
been found that the heating power of the same quan- 
tity of oak will be about 80, and of maple about 50. 
The practical value, however, of fuel may be governed 
largely by other circumstances. Thus, when a tire is 
wanted for only a short time, as a kitchen fire in mid- 
summer, or where steam must be raised quickly, etc., 
the cheaper fuel may be quite as valuable as the 
more costly. • 

Pages 78, 79. 

Arts. 122-125. — Momentum, according to New- 
ton's definition, is strictly quantity of motion. lie 
says (P/'incipia B. 1, Def. II.) " The whole motion 
is the sum of all its parts, and therefore in a body, 
double in quantity, with equal velocity, the motion is 
double; with twice the velocity it is quadruple." It 
is said in the text that quantity of motion does not 
fully express the desired meaning, but this is due sim- 
ply to the fact that quantity had not been defined. 
Including Newton's definition, it does express the, 
meaning correctly and fully. 

If a force of constant intensity acts upon a free 
body moving from rest, the product of the force and 



80 KEY AND SUPPLEMENT 

time equals the momentum produced. Space is here 
entirely abstracted from force and time. Although 
the body cannot move without involving space, ^yet 
all considerations of space must be discarded. It is 
immaterial whether the space over which the body 
must move in acquiring the velocity v be great or 
small ;, and hence, so far as the momentum is con- 
cerned, the space may vanish. 

From the equation Ft — Mv, Ave have F — M-, 
where —(or for a variable force, we have in the nota- 
tion of the calculus — ] is the rate of change of the 

velocity, and hence, M — is the rate of change of mo- 
mentum, which is, according to the second law, the 
measure of the force F. "We thus reproduce the ex- 
pression for that law. 

The expression Ft is not the momentum, but sim- 
ply its equal under the restrictions given above. It 
has been proposed to give a special name to this prod- 
uct, just as Miscalled the measure of work, while 
its equal Jil/t' 2 , in case of a free body, is called energy. 
Maxwell called Ft an impulse (Matter ami Ijfotion, 
p. 44), to which we do not object, since the effect will 
be the same whether it be produced in an impercep- 
tibly short time, or in a longer time. But the prod- 
. uct has no meaning except where the force moves 
a body. If F be a mere stress, like the pressure of a 
stone upon the earth where no motion is involved, 
then Ft produces nothing. The time effect of a 



TO ELEMENTARY MECHANICS. gl 

stress, is its effect in moving a body — producing 
velocity. 

Much has been written in regard to force, vis viva 
(now called energy) and momentum. .For many years, 
in the earlier history of the science of mechanics, there 
were long and sharp discussions as to whether work 
or momentum w T as the proper measure of force ; but, 
as we have shown in what precedes, neither is the 
proper measure, and hence theirs was merely a war of 
words. One factor in the product Fs is the force ; 
and one factor in the product Ft is also force. The 
former placed equal to \Mv % , and solved gives 

F — 2 - ; hence the measure of force is the rate of 

s ' J 

change of energy per unit of space; and we have 
already shown that it is also the rate of change of mo- 
mentum per unit of time. 

Efforts are sometimes made to determine a relation 
between momentum and energy ; but no physical re- 
lation exists, and hence none can be found. In order 
that there shall be a ratio between them, they must 
have a common unit. Since one is compounded of 
force and time in which space is excluded, and the 
other of force and space in which time is excluded, 
they have not a common unit. 

Page 79. 

We here note the three following statements from 
different authors. 

First, In Yan Nostrand's Engineering Magazine for 
1877 and 1878 is a series of articles on momentum, 
and Vis Viva by Prof. J. J. Skinner. On pp. 129, 



82 KEY AND SUPPLEMENT 

130, 131, it is stated that momentum which equals 
MV represents the number of pounds pressure 
which the mass M with the velocity V is capable of 
exerting under the conditions that the pressure is con- 
stant and capable of bringing the body to rest in one 
second. This is numerically correct as a deduction, 
but in the articles referred to there is apparently a 
labored effort to show that momentum is pressure 
only, and not quantity of motion (see also p. 137 of 
the Eng. Mag.). Also on page 132 it is stated that, 
" The unit of momentum, then, is a force of pressure 
equal to one pound " (see also p. 210). In this defi- 
nition, the element of time does not appear, but it is 
not proper to drop it simply because it is one second. 
The above-named writer corrected himself in a later 
article, page 501 of the same Magazine. 

To measure anything requires a unit of the same 
kind considered as a standard. Strictly speaking, the 
unit of momentum is the momentum of a unit of mass 
moving with a unit velocity. Momentum cannot he 
measured by pounds only. (See also article by the 
author, Eng. Mag., vol. xviii., 1878, p. 33.) 

Second, It is stated that Beaufoy determined that 
a body of one pound weight, with a velocity of one 
foot in a second, strikes with a pressure equal to 
0-5003 lb. ; and hence to find the pressure produced 
by the impact of any projectile, we have the general 
formula, pressure = 0-5003 TFV (Silliman's Physics). 
Now we assert that the formula is false. Admitting, 
for the sake of the argument, that he did find such a 
result when the projectile struck a hard body, like a 
piece of iron, it would have been very much less 



TO ELEMENTARY MECHANICS. 83 

had the body struck been more yielding, like a gas- 
bag, or a sack of loose feathers, and so a great range of 
values might be found depending upon the character 
of the bodies. 

Third, Professor Tait, in an interesting lecture upon 
force, delivered before the British Association, 1876, 
sa} T s (see Nature, 1876, p. 462), " With a moderate 
exertion you can raise a hundred weight a few feet, 
and in its descent it might he employed to drive ma- 
chinery, or to do some other species of work. But tug 
as you please at a ton, you will not be able to lift it ; 
and, therefore, after all your exertion, it will not be 
capable of doing any work in descending again. 

" Thus, it appears, that force is a mere name, and 
that the product of the force into the displacement of 
its point of application has an objective existence." 

" Force is the rate at which an agent does work per 
unit of length." . . . 

These definitions have already been referred to on 
p. 20 of the Key, and the remarks there made will be 
more readily understood in this place, after having 
passed over energy and momentum. Force, funda- 
mentally, is a quantity instead of rate • just as inter- 
est is a quantity — an amount of money — and not rate. 
It is true that the amount of money paid for the use 
of a hundred dollars is identical with the rate per 
cent., but every one readily distinguishes between 
rate per cent, and interest. Professor Tait made use 
of interest in an illustration of this principle, but, we 
think, used it improperly. 



8t KEY AND SUPPLEMENT 

Page 83. 

Art. 131.— If X = I, and k = 1, we have E = F, 

hence — as a deduction — the coefficient of elasticity 
may be denned as the force (stress) necessary to 
elongate a prismatic bar whose section is unity to 
double its length, provided the original conditions 
remain constant except that of length ; that is, the 
elasticity and the cross section must both remain con- 
stant. But these conditions are never realized, hence 
this definition is highly ideal. In fact, the coefficient 
of elasticity is constant for any material for an elonga- 
tion of only a very small fraction of the length ; but 
even with this limitation it is of untold importance 
in certain physical sciences. 
Page 90. 

Art. 138. — Substituting from equations (1) and (2), 
page 89, we have 

Mv? + M'v? = 

nr rMv + M'v' eM' . ,,-|2 

|_ M + M' M + M v ] J 

Mv? = /ly J/ ,,„ [MS? + IMM'm' + M' 2 c' 2 

1 (M + MJ[_ 

- 2eMM'v 2 - 2eM' 2 vv' + 2eMMvv' + 2eM' 2 

v' 2 4- e 2 M' 2 v 2 - 2e*M' 2 vv' + e 2 M'-v 2 l ; 

or 

= 1 \MH 2 + 2MWW + MM' 2 v 2 

(M 4- M'f L 



TO ELEMENTARY MECHANICS. 85 

- 2eM 2 M'tf — 2eMM' 2 vv' + 2eM 2 M'vv' + 
2eMMH' 2 + (?MM'H l - 2(?MM' 2 vv' + #M 

M' 2 v 2 \ (2) 

Similarly, 

M'v\ 2 = flkr M ' r , x , \MH 2 + 2MM'vv r + M" V 2 
+ 2eM 2 v 2 + 2eMM'm - 2eM 2 vv' - 2eMM' 
v' 2 + <s*af V - 2e*M W+ <?Mh 

= ^_L [~jf WV + 2MM" 2 W + JT 3 
( Jf + Jff ) 2 L 

a' 2 2eM 2 M'v 2 + 2eMM' 2 vv' - 2eM 2 M'w' - 
2eMM' 2 v' 2 + JlPM'v 2 - 2e 2 M 2 M'vv' + e 2 M 2 

JfV 2 ]. (3) 

Adding equations (2) and (3) we have 

Mv 2 + MV = {M ^ M J \MH 2 - 2eM 2 M'v 2 

+ JMM'H 2 + M*M' v 2 + 2eM 2 M'v 2 + e 2 M 2 
M'v 2 + 2M 2 M'm' - 2eMM' 2 vv' + 2eM 2 M' 
vo' - 2e L MM' 2 m' + 2MM'Hv' + 2eMM' 2 m' 

- 2eM 2 M'vv' - 2e*M 2 3f'vv' + MM' 2 v' 2 + 
2eMM l2 v' 2 + <?MM' % T) 2 + M'V - 2eMM , V 

+ e*M 2 M'v' 2 ~\ 



gg KEY AND SUPPLEMENT 



A__|~(Jf + M')MH 2 + 



(J/ + M')JMM'tf + (J/ + M') 2J/J/W 

(Jf + 3r)2^ 2 i!.0/W + (J/+ 31) M' 2 v* + 



(j/ +ifcr> 2 JOf , «' s l 



= _J___f JfV + e 2 MM'v 2 + 
2MM'vv' - 2e*MM'vv' + M*t>* + 6*JOf' 

= * -—S.MW + J[f V 2 + JO/ 7 

(eV + 2vv' — 2e 5 W + <?v' 2 ) I. 

Adding and subtracting 3IJI'(v 2 -f- ^' 2 j we have : 

+ M')3£'v 2 + MM\th* -v 2 + 2jw'(1 - <?)+ 

eV 2 — v' 2 ) 

= Mv 2 + M'v' 2 + J^,' [~(1 - 
M + M' L 

e ! )(2w'-« 2 -^)l 

M+ M 



TO ELEMENTARY MECHANICS. 87 

solutions of examples. 

Page 91. 

1. 5 x 5 = 25 lbs. sec. 



, Fl 9000 x 10 x 12 ft n . KQQ , . , 
2 ' A = 27T = 1 x 26000000 a0 ° 538 + mcheS - 



3. J£ = _^ = 2500x2x12 = 24j ^ 200 lb , 



very nearly. 



^ - TTV _ 120 - 120 
4 * K TT+ W ~ 18 



5. Eqs. (1) and (2), p. 89 of text, give 



Wv + W'v' eW , ,. 



20 x 100 + 50 x 40 | x 50 

70 70 

feet; 



(60) = 35* 



, 20 x 100 + 50 x 40 -x 20 CA 

«i = — ■ + ^— — x 60 = 

1 70 70 

65\ feet. 

Page 92. 

6. Here v' — 0, and we have 

2000 25 1AA n . f , 

v, = -yy - — x 100 = - 7\ feet, 



KEY AND SUPPLEMENT 



2000 + 10 xl00;=42|feet 
70 70 



7. Here v' = — 40 feet, hence 



2000 - 2000 _ j x 50 x 140 __ 
: W 70 

feet per second. 



50 



, 2000 - 2000 i x 20 x 40 _ on ,, 
Vi= _ .+ 75---^ «■ 

per second. 

8. By Art. 132, p. 84, we have 
9 It 

6 =v 

where h is the height of fall, and h' the height 
of rebound. The sum of an infinite decreas- 
ing progression is 

_ first term 
1 — ratio 

1-6* 

^ _h + he 2 






TO ELEMENTARY MECHANICS. 89 

9. If e = l, s = oo. 

e = |, s = \h 
e = I, s = \lh 

e = 0, s — It. 

10. Let M be the mass of one of the bodies, and v 
the velocity of the impinging one ; its kinetic 
energy will be 

The kinetic energy of the two bodies after im- 
pact will be least when both are non-elastic ; 
in which case the common velocity after im- 
pact will be (Eq. (1) p. 86), 

T=iv; 

and the kinetic energy will be 

and the kinetic energy before impact will be 
just twice that after, and cannot exceed that 
ratio. 

ANSWERS TO EXERCISES. 

1. Yes. 

2. No. 

3. No. 

4. By equation of work = energy. 

5. They will, at the instant of impact. 

6. Yes' 

7. No. 

8. No. 



90 KEY AND SUPPLEMENT 

9. By preventing so great a loss of energy. When 
the wheels of a car strike " dead " against a 
rail, it batters the rail, thereby doing an 
amount of work which is lost to the train ; but 
the springs, when in action, prevent such a 
dead blow in the first place, and then by their 
reaction restore a portion of the energy to the 
train. In short, if all the parts were perfectly 
elastic there would be no permanent battering 
of the parts, and no energy would be lost by 
the impact. 
Page 94. 

Statics is a limiting case of dynamics, in which the 
applied forces mutually destroy each other, and leave 
the bo^dy, so far as its condition in regard to rest or 
motion is concerned, the same as if no forces were 
acting. As such, its principles may be established in- 
dependently of motion. Many writers hold that the 
conditions of equilibrium should be determined inde- 
pendently of all considerations of motion, and we 
have accordingly given the usual proofs, although 
the Newtonian method, given in Art, 52 of the text 
is, to us, quite satisfactory. 

We consider that the confirmation of the results, 
flowing from the parallelogram of forces, is a stronger 
confirmation of its truth than any formal demon, 
stration ever made. Nearly all the formulas of me- 
chanics are founded upon it. The principles of mech- 
anism, and the places of planets and comets all 
involve it. Such a proposition might be assumed 
without proof, and its truthfulness be confirmed by 
its leading to results confirmed by daily experience. 



TO ELEMENTARY MECHANICS. 91 

Formerly a proof was considered so important, that 
many different methods were devised, and one work 
gave forty-five different proofs of the parallelogram 
of forces. 



solutions of examples. 

Page 99. 

1. Let P be one stress and J^the other, then we 

have (Eq. on p. 97 of text), 

B = VP 2 ~T~F~~T~2PF cos 90° = VP 2 + F 2 . 
If B ■= 0, then 

R = VP 2 + * 2 + WF = P + F. 

If 6 = 180°, then 

i? - VP 2 + ^ 2 - 2PF = P - F. 

2. Here the force 5 reversed will be the resultant 

of the other two ; then 

5 2 =3 2 4- 4 2 + 2.3.4 cos 0; 
25-9-16 



24 

1 = 90°. 



= 0: 



3. We have 



P 2 = P* + P 2 + 2BP cos 0', 

.'. cos — — ^ • 
.-. ^ = 120°. 



92 KEY AND SUPPLEMENT 

4. We have 

P 2 = 100 2 + 100 2 + 2 x 100 x 100 cos 60° 
But cos 60° = |, hence 
P 2 = 3 x 100 2 ; 
.-. R = 100 V3. 

5 "We have 

(P + Ff = F> + F 2 + 2PFcos 0; 
or, 
P 2 + 2PP + P 2 = P 2 + P 2 + 2PP cos 0; 

.'. cos 0=1, 

and 

(9 = 0°. 

6. We have 

B* = (P- Ff = P 2 + P 2 + 2PP cos 0, 
or, 

P 2 - 2PP+P 2 = P 2 + i^ 2 + 2PP cos 0; 
.;. cos = - 1 ; 
.-. 0= ISO 3 . 

7. From the proportion on p. 9S of text we have 

50 : F:: sin (P,P) : sin 115°, 
50 : P :: sin (PP) : sin 35°, 



TO ELEMENTARY MECHANICS. 93 

F: E : : sin 115° : sin 35°, 
P : E : : sin 30° : sin 35°. 

From these we have 

sin 3o 

B = ^J^ o x 50 = 57-35 lbs. 

sin 30 

.-. F = 90-63 lbs. 

sin (^,7?) = 5 °g 7 f 3 n 6 85 = 0-5000 ; 
.*. angle (F,IZ) = 30°. 

8. The string will make a right angle at the point 

where the weight is applied, and the sides of 
the triangle representing the forces will be as 
3 to 4 to 5 ; hence we have 

5:4 ::20 :x = 16; 

5 : 3 : : 20 : % = 12. 
Page 100. 

9. The parallelogram representing the forces will 

be a rectangle, of which the diagonal will be 
a diameter of the circle. 



ANSWERS TO EXERCISES. 

1. It will be the resultant of two forces acting 
ajvay from C, one of which will equal CA, the 
other CD = AB. 



94 KEY AND SUPPLEMENT 

2. A line through A equal and parallel to a line 

joining .2? and B. 
3 They would not. 

4. When acting upon the same particle, in opposite 

directions, and equal in magnitude. 

5. With 4, 5, and 9 they can, if 4 and 5 act opposite 

to 9. But forces 3, 4, and 8 cannot, since two 
of them, 3 and 4 together, do not equal the 
third. 

6. It will not. The resultant takes the place of 

the other two. 

Page 102. 

Art. 158. It will be observed that, in Fig. 45, /3 
is the complement of a ; hence cos /3 = cos 
(90 — a) — sin a. hence the equations for Xand 
Y become 

X = F x cos a\ + F 2 cos a 2 + etc. = 0, 
Y= JF X sin a 2 + F % sin a 2 + etc. = ; 

from which we see that the equation for Y 
may be deduced directly from that of X by 
writing sin in place of cos in the first equa- 
tion. 

solutions of examples. 

Page 103. 

1. We have 

X =z 20 cos 30° + 30 cos 90° + 40 cos 150° 
50 cos 180° = R cos a, 



TO ELEMENTARY MECHANICS. 95 

T= 20 sin 30° + 30 sin 90°+ 40 sin 150° + 50 

sin 180° = R sin a • 

in which all the terms are written as positive, 
and their essential signs made to depend upon 
the trigonometrical functions. 
Reducing gives 

20 x iV3 + 0-40 x |V3 - 50 = R cos a, 
10 4- 30 + 20 + = 72 sin a; 





- 67-32 + = R cos a, 




60 = R sin a. 


Dividing gives 




.7? cos a 

R sina 


67-32 

60 ' 



cota= -1-122; 
.*. a = 138° 17'. 

Squaring and adding gives 

72 2 (cos 2 a + sin 2 a) = (67-32) 2 + (60) 2 . 
But cos 2 a + sin 2 a — 1 ; 



72 = V8131-98 
= 90-18 lbs. 



96 KEY AND SUPPLEMENT 

2. R cos a = 20 cos 180° + 10 cos 270°, 

E sin a = 20 sin 180° + 10 sin 270° ; 

hence 

P cos a— — 20, 
7? sin a = — 10. 

Squaring and adding gives 
i2= V500 = 22-36. 

3. We have 

P cosa = P cos 0° + P cos 90° + P cos 225° 

+ P cos 270°, 
R sin « = P sin 0° + P sin 90° + P sin 225° 

+ P sin 270° ; 
hence 

R cos a ---- P (1 - |a/2 ), 
Rsina =P(1 -iV2-l) 
= - i V2P 

Dividing the second by the first gives 

-24162 + 
.-. a = 292° 30'. 



TO ELEMENTARY MECHANICS. 97 

Squaring and adding gives 

_# 2 (sin 2 a + cos 2 a) = P*(l - I V 2) 2 + P 2 (- -|V2) 2 5 
or 

7? 2 = I»(2 - V2) 

= 0-5S579P 2 ; 
.-. #=0.765P. 



Page 104, Art, 162. — A single force whose line of 
action does not pass through the centre of a free 
body, produces rotation as well as translation. The 

. measure of the effect of a force in producing rotation 
is proportional to the moment of the force; as is 
shown from the fact that the moment is proportional 
to the work done oy the force. 

The theory of moments is here discussed without 
reference to the bodies upon which the forces act. 

Page 110, Art. 176. — The cut, Fig. 55, should be as 
here given ; that is, in the 
typical figure the force should 
be positive away from the 
origin, and so placed that it 
would produce positive rota- 
tion about the origin, O, and 
the angles a and f3 be acute, 
as shown in Fig. 42 of the 
text ; so that the signs of the terms in the analytical 
expression for the moment will flow directly from 
the trigonometrical functions. This being done the 
5 




98 KEY AXT> SUPPLEMENT 

typical form of the expression for the arm of the 
force will be 

Oa = x cos fi -y cos a. 

In the old Fig. 55, the' angle between the axis of a? 
and the direction line of the force was, as shown 

in Art. 157, 

a = 180° + dOb ; 

and similarly, 

ft = 180° + YOcl ; 
.-. cos a = — cos dOb, 
cos fi = — cos l r 0d, 
which values give for the arm 

Oa — ycosa — x cos /?, 

as given in the text. We call the former value the 
typical one, and the latter a deduced one. The for- 
mer should always be used in connection with the true 
value of the angles a and §. 



SOLUTIONS OF EXAMPLES. 



Page 115. 

1. We have 



1 - IE w > 

in which 

TF= 20 lbs., 

AC = 24 inches, AB = 6 inches, AD = 4 
inches. 



TO ELEMENTARY MECHANICS. 99 

To find AE ; in tlie right-angled triangle DAB 
we have 

tang B = ^g = i = 0-666 + ; . 

■\ B = 33° 41'. 
Then 

AE = AB sin B 

= 3-327 + inches. 

Substituting above gives 

t= "3^2T = W4 ' 2 + lbS * 



2. Taking the origin of moments at D we have 

AE V 
= 20x21 = 1201bg> 



3. Taking the origin of moments at B we have 
W.BO=\F.AB, 



i^ 2 ±^601b, 



Page 116. 

4. If t = W. we have 



100 KEY AND SUPPLEMENT 

VV AE n > 



AE = AC. 
From the right-angled triangle AEB we have 

sin -io 
substituting, 

_ AC 

. ~Vi 
= V2AC. 

5. Taking the origin of moments at A we have 

t x AC=BEx W. 
But from the example 
DB = 2AB,0 = 45°, TF= 50 lbs. 
From the figure 

BE=ABsm4:5 

= iV2AB = 0.l071AB. 

BE AB sin 45° 
sin cp = — =—- _ = _ — . 

.-. <p. = 20° 42' 17 . 
To find .4(7 we have from the figure 

AC- AB win ABC 

= AB sin (0 - cp) 



TO ELEMENTARY MECHANICS. 101 

= AB sin 24° 17' 43" 
= 04115 AB. 

Substituting in the first equation above gives 

0-7071 x 50 



fc=- 



04115 

85-97 + lbs. 



To get the compression on the bar, take the origin 
of moments at D, in which case the moment of 
the tension will be zero. The perpendicular 
from D upon AB prolonged will be AD sin 6, 
and from D perpendicular upon the vertical 
through B will equal BE = AB sin 6 ; hence 
we have — calling c the compression — 

. c.AD&m6= W. AB sin 0; 

. . e - AD W. 

To find AD we have 
sin ABD : sin cp : : AD : AB ; 

• AD - AB x °' 4115 

AB m 
0-8592 + ' • 

which substituted above gives 

c = 0-8592 W > 
= 42'96 lbs. 



102 KEY AND SUPPLEMENT 

Page 116. * 

6. Let fall a perpendicular^ from B upon AC, then 

p = AB sin A. 
Let 1) be directly under W, then taking the 
origin of moments at B we have 
fTp= W.BD ; 

. BC cos CBD ^ 
AB sin yi 

7. If * = IF,' then AB sin j^tf = ^7>, or BC will 

bisect the angle ACD. 

8. Take the origin of moments at ^. Let fall a 

perpendicular from A upon CB produced; its 
length will be 

<p = AB sin CBD 

Q 

= 6 x 



V± 2 + 8 2 

48 



feet. 



~~ 8-9442 

Let c be the compression on BC, then the 
equation of moments becomes, 

c.p= W.AD; 

/. c = W 



p 

10 x 8-9442 

48 . 
: 931-7 lbs. 



500 



TO ELEMENTARY MECHANICS. 103 

9. For equilibrium we have 







~ Fi- 


-p 1 ■- 






_2P t 










= go; 






or there can 


be no < 


equilibrium 


10. 


We will hav 


e 





be = P °~ Pl Ob 

_ 2 x 
~ Pi 
= 0; 

or the forces must act at the same point. 
Page 118, Art. 187. — It is well to illustrate this article 
still further. If the forces IP 

constituting the couple be A c j ? B 

equidistant from the cen- ct * 

tre g of the body, it is suffi- |p 

ciently evident, without a thorough demonstration, 
that it will produce rotation only. But is it equally 
evident that, if the same couple act upon the same 
body in such a w T ay that the ,p 

points of application are 

both on one side of the cen- ^ ■•-^ £? 

tre, it will produce rotation 
only, and the same amount 
of rotation as in the preceding case ? According to 



104 



KEY AND SUPPLEMENT 



-9 



HP 



HP I 



HP 



61 
HP I 



the proposition it will, and we will prove it by a special 
solution. 

At a point d, such that do = cb, introduce two 

equal a n d 
opposite 

forces, each 

equal to|P, 
and call the 

upper |P, 1, and the lower half 4. Since these are 
in equilibrium the problem will be the same as before. 
Separate the force at b into two equal parts, and for 
the sake of convenience call one part 2 and the 
other 5. Now the resultant of 4 and 5 will be a 
force equal to their sum, or P, applied at the centre 
c — which force call 9. Combining 1 with 2 we have 
a. couple whose arm is db, and the moment will be 
\P.db = P.cb. Similarly at <?, at a distance ec = 
ca, introduce two equal and opposite forces, each 
equal to \P ; and separate the force at a into two 
equal parts. Combining 4P at a with the \P above 
e gives a resultant equal to P applied at c acting- 
down, and marked — 9, which will equilibrate + 9, 
and there will be no motion of translation. There 
will remain the couple \P.ea, which will produce 
rotation only about the centre c. Finally, the two 
couples \P.db, and }P.ea, ench producing rotation 
about the centre c, but in opposite senses, arc equiva- 
lent to the single couple. 

P(idb - \ea) = P{cb - cci) = P.al ; 

hence, the body will rotate about its centre of gravity, 



TO ELEMENTARY MECHANICS. 105 

and the rate of rotation will be independent of the 
points of application of the forces, and dependent only 
upon the moment of the couple. 

Page 120, Art. 190. — This article contains all the prin- 
ciples of the simple lever. In some works levers are 
divided into three classes, but, mechanically, there is 
no distinction between them. It is only necessary 
for equilibrium that the sum of the moments be zero 
— or that the moments of the forces which turn the 
lever one way equals the moments of those which 
tend to turn it the opposite way. 

Page 121. 

Art. 192. — The force F' and D produces equilib- 
rium in the system ; hence a single force equal and 
opposite to F at D will produce the same effect as 
the three forces F, F, and P. 

Art. 197.— Assume that any number of forces act 
upon a body in any manner; they may produce both 
translation and rotation. The measure of their effort 
to produce rotation will be the sum of their moments, 
wherever be the origin, and the sum of these mo- 
ments will be equivalent to a single couple, Arts. 184 
and 186. Hence, if there be no effort at rotation, the 
sum of the moments will vanish for any and every 
point assumed for the origin of moments. The only 
other tendency to motion is that'of translation; in 
which case there will be a single resultant passing 
through the centre of the body. If there be a result- 
ant, the moment will be zero when the origin of mo- 
ments is on the line of the resultant, Art. 189 ; and 
will have a finite value when the origin is not on the 



10(5 KEY AND SUPPLEMENT 

line of the resultant ; and if tliere be no resultant the 
latter moment will also vanish. 

solutions of examples. 
Page 125. 

1. For, according to the triangle of forces, the re- 
sultant of two of them will equal in magnitude that 
represented by the third side, but its direction of ac- 
tion will be opposite to that represented by the third 
side, and at a distance from it equal to the altitude of 
the triangle. See Fig. 71 of the text, only in this 
case the third force will act from A towards B. 

2. Inscribe a circle in the triangle ; then will the 
radius r be the common arm of the three forces in 
reference to the centre of the circle, and we will have 
the equation of moments 

B.r = P.r + F.r ; 
.:fi = P+F. 

3. The point of application of the resultant of two 
of them will be at the middle point of the side of the 
triangle between them ; and the point of application 
of this resultant and the third weight will be at two- 
thirds the distance from the third weight on the line 
joining them ; and this will be the required point. It 
is at the intersection of the medians of the triangle 
(Art. 222). 

4. Let P and Pbe the respective amounts; then 
taking the origin of moments at the weight we have 
the relative moments %P and l.F\ hence 

2P = F. 
Also, since the sum of P and F equals the entire 
weight. 



TO ELEMENTARY MECHANICS. 107 

P + F= 175. 

Substituting, 

ZP = 175 pounds ; 
.*. P = 5§} pounds, 
- and 

F = 116| pounds. 

5. Let W = 500, the arm of which in reference 
to the point B will be one-half of AB, or 1 foot ; 
and the arm of K will be DB — 3 feet ; hence the 
equation of moments will be 

3i^=lTF 

= 500 lbs. ; 
.-. i^=166|lbs. 

6. Let x = the required distance, then will the lever 
arm of the weight be x, and of the man 8 — x ; hence 
we have, taking the origin of moments at the fulcrum, 

175 (S - x) = 4000a?, 
or ' 

(4000 + 175)x = 8 x 175; 
r.x=m%feet 
= ^ryf-y inches. 

answers to exercises. 
Page 126. 

1. It is. Foot-pounds of rotary effort. 

2. The resistance in pounds which is overcome 

through a certain number of linear feet. 

3. It is the momentum of a given number of 



108 KEY AND SUPPLEMENT 

pounds of mass moving at a given rate in feet 
per second. 

4. This question is defective, because velocity in- 

volves a unit of time, which may be one sec- 
ond, one minute, or any other unit. Assum- 
ing that the velocity is feet per second, the 
unit will be 1 pound of mass x 1 foot per 
second x 1 foot for the arm. 

5. It can, and will always <k> so in a free body if 

the line of action of the force does not pass 
through the centre of the mass. 

(5. They cannot. Since neither couple acting sepa- 
rately can produce translation, they cannot 
produce it when acting together. The resultant 
of two couples is a single couple, Arts. 185 and 
186, and for this reason can produce rotation 
only. 

7. It will be 100 lbs. more. Pulling down with his 

hands will add nothing to the pressure of his 
feet, for that effort is resisted by an equal up- 
ward push of his shoulder. A man by pulling 
upward on the straps of his boots, does not, 
thereby, diminish the pressure of the boot 
upon the floor, although it increases the pressure 
between his foot and the boot. No u perpet- 
ual motion man" has yet thrown himself over 
a fence by pulling on the straps of his boots. 

8. If the cutting is uniform it is ; but if the timber 

at one side of the hole is harder than at the 



TO ELEMENTARY MECHANICS. 109 

other, there will be a side push, tending to 
force the auger out of line. 

9. It will. 

solutions of examples. 
Page 133. 

1. Let R be the resultant, and x the distance of its 

point of application from the force 6. Take 
the origin moments at 6 ; then we have the 
equation of moments 

Rx = 11 x 5, 

and of forces 

R=6 + 11 = 17; 

.*. x == yy ' = ^- Oj y ieei. 

2. "We have, using the same notation as before, 

Rx = 11 x 5 = 55, 
R = 11 - 6 = 5 ; 

.-. x — - 5 f- = 11 feet. 

3. Take the origin of moments at the extremity of 

the line near the weight 2, and retaining the 
same notation as above, we have 

Rx = 2-2 + 3-3 + 4-4 4- 5-5 = 54 
5=2 + 3+4+ 5 = 14 lbs.; 

.-. x = 54 + 14 = 3f feet. 

4. Take the origin of moments at A, then we have 



HO KEY AND SUPPLEMENT 

i?z =3x0 + 4x3 + 5x7 + 6x5 = 77; 
R = 3 + A + 5 + 6 = 18 lbs.; 
/. aj = 77 -T- 18 = 4 feet 3^ inches. 
Page 134. 

5. Let as be distance from A where P is applied, 
^to the point of application of the resultant; 
take the origin of moments at the point of ap- 
plication of the resultant, then will the arm 
of P be x sin <p,and of F,(x + AB) sin <£/, as- 
suming P > F; hence 
Px sin cp = F{x + AB) sin <p ; 

:.x =j r -—yAB; 

which substituted in the preceding equation 
gives 

PF . PF 

?. ri AB sin w — t5 T , AB sin ro. 

P — F ^ P — F 

answers to exercises. 
Page 134. 

1. It has not. We may say that it has one at in- 

finity, which is equivalent to saying that it has 
none. 

2. Iso. The sum of the forces must also be zero. 

If the sum of the moments in reference to 
three arbitrary points is zero, they will be in 
equilibrium (see Art. 195 of the text). 

3. When they form a couple. 

4. Because the resultant is zero. 

5. It will. 

6. It will. 



TO ELEMENTARY MECHANICS. m 

7. It may vary directly as the distance from the 
centre; or inversely as the distance ; or as any 
power of the distance ; or any root of the dis- 
tance ; or as any power for a part of the dis- 
tance, and any root for the remaining distance ; 
or in any other way, provided the concentric 
shells comprising the sphere shall be of uni- 
form density. 
Page 135. — Many of the properties of the centre of 
gravity were developed as long ago as in the days of 
x\rchimedes. The properties are not only of great 
importance in statics ; but when the principles of 
dynamics were developed — after Galileo's time — they 
were found to be no less important in that science. 
We mention only one — A free rotating body rotates 
about an axis through its centre of gravity ; or, more 
strictly, through the centre of the mass. 

answers to exercises. 
Page 138. 

1. The vertical through the centre of gravity of 

the carriage and load must intersect the ground 
between the wheels. In order that it shall over- 
turn, the vertical must fall outside the base of 
the wheels. A carriage in motion may. over- 
turn when it would not if standing, on account 
of the inertia of the mass. A sudden side- 
"lurchj' may induce a rotary movement suffi- 
cient to overturn it. ~ - 

2. He can stand so long as the vertical through the 

centre of gravity of his body falls within the 
base occupied by his feet. 



112 KEY AND SUPPLEMENT 

3. Some parts of his body must move backward. 

The space occupied by hfs feet being of Unite 
size, he may move his head, or other parts of 
his body, to some extent without endangering 
his stability. 

4. Because the base being so very narrow, with only 

two legs, a small displacement will cause the 
vertical through the centre of gravity to fall 
without the support. 

5. Because in Fig. 77 the centre of gravity must 

be moved further than in Fig. 78, and also 
must be raised through a greater height. Some 
writers consider the height through which the 
centre of gravity must be raised in order to 
overturn a body, a measure of its stability. 

6. Because the line through the point of support 

and the centre of gravity of the book will be 
inclined to the edges, and as the former will 
be vertical, the latter must be inclined. 

7. It may. Such will be the case with a cylinder 

resting on its convex surface. It will be in 

indifferent equilibrium in reference to rolling, 

but stable in reference to a longitudinal mo- 
tion. 

8. The centre of gravity is below the top of the 

post. 

9. In both these exercises the balls are so much 

heavier than the bodies to which they are at- 
tached, that the centre of gravity of the whole 
device is below the support. 



TO ELEMENTARY MECHANICS. H3 



Page 140. 

x is read " x dash" 



solutions of examples. 

Page 140. 

1. Let I be the length of the line, and x the dis- 
tance of the centre of gravity to the smaller 
weight ; then will I — x be the distance to the 
other, and the equation of moments gives 

1.05 = n(l - x) ; 

n 



I 



1 

If n = 2, then x = \l. 
If n = 3, then x — \l. 

2. The middle of one side will be the centre of 

gravity of two of them, and the centre of 
gravity of the three will be in the line joining 
this point with the vertex, and, according to 
the preceding example, it will be at two-thirds 
the distance from the apex. Hence the centre 
of gravity will be at two-thirds the distance 
from any apex, on a line drawn to the middle 
of the opposite side. 

3. The centre of gravity of the weights 1 and 

2 will be in the line joining them, and at 
two-thirds the distance from 1 ; and of the 
three weights it will be in the line joining the 
former point, and the weight 3, and at its 



114 KEY AND SUPPLEMENT 

middle point. The solution is the same whether 
the triangle be equilateral or scalene. 
4. The centre of gravity of the three weights at 
the base will be at the centre of the base, and 
of the four weights in a line joining the apex 
with the centre of the base, and, according to 
the first example, at three-fourths the distance 
from the apex. 
Art. 217. — A line, in mechanics, is a body from 
which all dimensions are abstracted except 
that of length. 
Page 142, Art. 220. — This article is introduced here — 
in advance of its proof — partly to classify it with 
lines, and partly to furnish exercises. 

SOLUTIONS OF EXAMPLES. 

1. Join the centre of gravity of one edge with that 

of another, and the centre of this line with 
the centre of another edge, and so on. 

2. Half the diagonal of the base (a side being 1) 

will be 1a/2, and the length of one of the lat- 



eral edges will be V'l + (| ^/2f - Vf = W§. 
There are four edges, and hence the entire 
lengths will be 2\/6. The lengths of the 
sides of the base will be 4. The centre of 
gravity of the four lateral edges will be at one- 
half the altitude, and of the edges of the base 
it will be at the centre of the base. Taking 
the origin of moments at the apex, and x the 
distance of the centre of gravity from the apex, 
we have 



TO ELEMENTARY MECHANICS. 115 

(2a/6 + 4) x = 2^6 x i + 4 x 1 , 

'.\x = o- , m +. 

3. In the equation in Article 220, make BO = r, 

AC = %r ; then arc ABC = nr, and we have 

Q ^_ r.2r __2r 
7tr n 
Page 143. 

4. We will have BO = r, AC = the chord of 

60° = r, the radius, and AB — %nr ^% = Jtzt ; 
hence 

/i _ r.r _ 3^ 
~ \nr ~ ?t' 

5. We will have BO = r, AC — the side of an in- 

scribed square == r^/2, and ABC = the arc of 
a quadrant = -|7T^ ; hence 

Akt. 221. — A surface is a body from which all di- 
mensions are abstracted except length and 
breadth. 

solutions of examples. 
Page 145. 

1. It will not, for the moment of the part next to 
the apex will be less than the part next to the 
base. 

' 2. When the sides adjacent to the vertical angle 
are equal, the bisecting line will pass through 



HQ KEY AND SUPPLEMENT 

the centre of gravity of the triangle. In other 
cases it will not. 

3. The altitudes will be the same, and the centre 

of gravity will be at one-third the altitude from 
the base. 

4. The area of the larger circle will be nil 2 ; of 

the smaller nr* ; and of the remaining part 
n {j& _ 7 -2). Taking the point A for the origin 
of moments, we have 

n(P - r 2 )Ac = 7t1l\R - nr\r ; 
7l 3 — v 3 
* ii! 2 -/- 2 ' 

_ R 2 + Rr + r 2 
U + r 
If r = Ji?,then 

Ac = ifi. 
Ifr= R, Ac ==-fff. 

5. Let AF — AD — a ; then the diagonal AE = 

aV%, CE — laV% and the equation of mo- 
ments will be 

\a 2 .AB = "A^V2 - ia 2 (aVZ - Ja a/2) ; 
.-. AB = &aV2 =H«V2 = $AC. 

Page 146. 

6. The centre of gravity of the triangle ABC will 

be at i of CFfYomF', ot DOE, \ of CG from 
G\ or from Fit will be FG + J- 676?. The tri- 
angle ^£6' will be to triangle DCE as (FCf 



TO ELEMENTARY MECHANICS. \yj 

is to (GO) 2 ; and ACB: ADEB = (FC) 2 : 
(FCf-{GCf. 

Taking the origin of moments at F we have, 
ADEB x Fg = ACB x ±FC - DCF x 

(FG + \CG). 
From the proportions, we have, 

ADEB = ACB J^- c \?^\ 

DCF = ACB x ^L. 

(FC-) 2 

Substituting, 

^FCf - (GCf)Fg = \(FCf - (CG)\FG + 

iCG). 
From the figure we have 

CG _DE 

FC ~ AB' 
and 

FC= CG-{-FG; 

■ CG - DE . 
*'' CG + FG ~ ^i£ ' 

hence we find 
Substituting above gives 



Hg KEY AND SUPPLEMENT 

Reducing gives 
(^5)' - (DEf __ r (^) 3 _ , 

(AB-DEf J 

r( ^g) 3 - SAB.jBEf + 2(DEY ~] 
_ij^— {AB-DEf 

.'. ity = i^ 

r (AB f - %AB(DEf + % DEY "1 
L("-A^ - DE)(AB - DE){AB + ZM')J 

_,„ r (^^) 3 - zAB.jD Ey + 2(i>ff) 3 i 

- ^ LC^^ 2 - 2AB.BE + DE 2 )(AB +1)E)A 
* ^ -JBT^E' 



7. The slant height will be \A' 2 + h 2 , and the lat- 
eral area, 27tr.IV> 2 + A 2 ; and the centre of 
gravity of the lateral area will be in the axis 
at § h from the apex. Let x be the required 
distance, then 



(tzt 2 + nr Vr 2 + A 2 ) x = nAh + TtrV? + h\\h ; 



— 3 Vr 2 + f$ 4- r 7 

.*. a? = — , h. 

V>* + h 2 + r 



TO ELEMENTARY MECHANICS. 



119 



To find the centre 




Page 149, Art, 230. — Problem. 
of gravity of a segment of a 
sphere. 

A segment of one base, GAIT, 
is considered in the text. 

To find the volume of the A / 
segment A GIF, we have from 
geometry 



±7z{GIIf + iGHx 7r(ABf 
= nGH{\GH* + i{ARf). 

We put this under another form — thus, in the right- 
angled triangle AJ7C, we have 

(AHf = (AC? - (ON? . 

= (GGf- (CG-GIlf 
= WG.GIT- (Gllf 
- = GH(2CG-GHy 

which, substituted, in the preceding expression, gives 
for the volume of the segment, 

7t{GH)\CG-lGH\ 

which is the value used in the text. If r be the radius 
of the sphere, and h the altitude of the segment, the 
expression becomes 



7th 2 (r-^h\. 

For the volume of the spherical sector ACG, we have, 
from geometry, 



120 KEY AND SUPPLEMENT 

2n(CGf x^xiCG 

= ^{GGf x GH 

= |w% 

which is also the value used in the text. 

The volume of the cone the radius of whose base is 
AH, and altitude HC, is 

7T{AIIf x \CI1 

= i7t(2r7i - h 2 )(r - h) 

= i7r7<(2r-h){r-l<). 

These values in the first equation of the article 
gives 

r , _ S(CG)\GTLCg - 3(AH)\(CIiy 
9 " 12(GH)\CG-±G1I) 

_ $(r\h.j(2r - It) - 3(;- 2 - (r - 7<) 2 )(r - If 
J2A 2 .(r-i/0 

= 1 (2/' ~ W 

To put this under another form, let the angle 
ACG = 0, then 

h ±z r — CH = r — r cos 6 = r(l — cos 6) ; 
hence 

2r — h = 2r - (r - r cos 6) 
— 2r — r 4- r cos 6 
= r (1 4- cos 6). 



TO ELEMENTARY MECHANICS. 121 

From trigonometry 

2 cos 2 |-0 = 1 + cos ; 
.-.(2r - l,f = 4=r 2 cos 4 id. 

Also we find 

r-iA = r-i(CG-OII) 
= r — ±(r — rco&6) 
= |7>(2 + cos 0) ; 

and these substituted above, give 
3cosHfl 

If A == 0, we have 0=0, and the segment will vanish, 
and we have 

ey = r , 

as it should. 

If h = %\ d = 180°, and we have 

Cg'=0 y 

hence the centre of gravity of the sphere will be at 
its geometrical centre, as it should. 

For the hemisphere, 6 = 90°, and we have 

= fr. 

If the segment has two bases, let 6 be the angle 
subtended by the radius of the upper base, and <p the 
angle subtended by the radius of the lower base, then 
by taking the difference of the moments of both seg- 
ments, and the segment on the upper base, we. find 



122 KEY AND SUPPLEMENT 

<v = 

(1 - cos cpf. cos'ly - ( 1 - cos Of. cos 4 *i9 
° (1 — cos cpf{'& + cos cp) — (1 — cos 6>) 2 (2 + cos 6) 

solutions of examples. 
Page 150. 

1. In Article 229, CG is the radius of a circle, and 

by the conditions of the problem GIT will also 
be a radius; hence we have 

%(WG - Gil) 

= |(2r-r) 

= !>•• 

2. Let x be the required distance to the centre from 

the common tangent point ; then, according to 
Article 202, the moment of the difference of 
the spheres will equal the difference of their 
moments, and we have 

(J ?rR 3 - ±ni*)v =a \7iR\R - \nr*r ; 

R" - , A 
•*• x = Jga_ r z 

B* + R 2 r + Rr + /- 3 



R 2 + 7fr + ** 

If ?• = 0, we have 

x = R, 

and the centre of gravity will be at the centre 

of the sphere. 

If r — i?, we have 

X = -3 XI, 

which gives the centre of gravity when the 



TO ELEMENTARY MECHANICS. 



123 



thickness of the solid, opposite the tangent 
point, is infinitesimal. 
3. To find the versed-sine Gil, Fig. 94, we have 
in this case OB — r, HB = \r ; 



Cg' 



.-. 011= vV - T y- 2 = J/V15 ; 

..-. GR=(l -iVl5)r. 
In the equation of Article 230, 
Cg = |[2r - (1 - iVl5)/-] = 1(1 + lVTB)r, 
and we have 
= 8/-(l-lVr5)/^I(l + iy/T5)/'-3(H 2 .(^Vi5) 2 

12(1 - WW x i* x [(r -i)(l - iyl5)r] 



12(B - iVl5)(i + T VV15) 
= 0-977 r nearly. 

A 




4. The centre of gravity will be J- of 8 inches, or 
2 inches from the base, hence the line joining 



124 KEY AND SUPPLEMENT 

the point of suspension with the centre of 
gravity forms a triangle of which the two sides 
are each 2, and as the altitude is perpendicular 
to the base, the oblique angles will each be 
45 degrees, which will equal the required in- 
clination. 

5. In this case the angle between the perpendicu- 
lar and the radius of the base will be' 30°, and 
we have 

tan 30° — \Alt. -± radius of base ; 

Alt. of cone A 

• • — 7 • * i = ^ tau 30 

radius of base 

iV'S V3 
= |V3 
= 2-30940+. 

solutions of examples. 
Page 154. 

1. In this case 6 — \it in the equation of Art. 235, 

and sin 6 = 1 ; hence, by substitution, we have 

<* = *£. j 

2. The area of the segment will equal the area of 

the sector ACBG, minus the area of the trian- 
gle ACB. Let be the angle ACG, then 
ACB = 20, arc AGB = r.26 = 2r0, and the 
area of the sector = }r.2r6 = r 2 6 ; HB = r 
sin 6, CH = r cos #, and area of the triangle 
ACB = 7* sin 6 cos 6. 



TO ELEMENTARY MECHANICS. 125 

Hence 
7 s (6 - sin 6 cos ff)x = *»0.|?^? - r 2 sin 6 

cos #.§/• cos 0; 

_ sin — sin cos 2 (9 

•'• « = 1 ' o - sin cos 6* " '' 

sin 3 



6/ — sin 6 cos 



3. The sphere may be generated by the revolution 

of a semicircle about a diameter. Area of the 
semicircle = \iti 2 . The circumference de- 
scribed by the centre of gravity of the semi- 
circle, will be (Ex.1), f- • 2tt — \r. Hence, 
according to Art. 233, we have 
volume = \ni*'\r — J | 7r/ ' 3 - 

4. Referring to the solution of Example 2 above, 

we hnd, 

area of segment = (6 — sin cos 0)/- 2 , 
which multiplied by 27tx, where x is the an- 
swer to the 2d Example, gives the required re- 
sult ; hence the required volume is 

• %7tr % sin 3 
= volume of the sphere x sin 3 0. 

5. Following the method of Art. 233, and using 

the values already found, Ave have 



126 KEY AND SUPPLEMENT 

volume = r 2 6 x 27rCg 

= 1 7T?' 3 Sill 

= volume of the sphere x sin 
= ft*r*. A£T 

— area of great circle x %AIL 

If = 90° we have *7T/ >3 , which is the volume 

of a sphere, as it should be. 
The volume generated as in Example 4 = vol. 

in Ex. 5 x sin 2 0. 

Page 158, Art. 241. — In the first edition of this work, 
the solution of this problem is erroneous. The line 
EF will not generally be tangent to the curve MN , 
and hence the analysis founded on that supposition is 
erroneous. They will be in equilibrium when the 
point g is vertically under C. 

Let EgC = <p, then the equation of moments will be 

W\,Eg .sin cp — W % .gF. sin cp ; 
.-.TFiJfr'.- W t .gF. 

From the figure we have 

Eg + gF=EF=r l + r„ 

which combined with the preceding equation gives 

also 

CF=B-r ii 
CE=B-r u 



TO ELEMENTARY MECHANICS. 127 

and as the three sides of the triangle EOF thus be- 
come known, the angles i^and i^may be found; for 
we have from trigonometry 



4 ab 

where s = \{a 4- h + c), a, b, c, being tlie sides of the 
triangle, and is 7 the angle opposite the side <?, or OK 
Similarly, 

be 

where i^is opposite a, or EO. 

In the triangle EOg the two sides CE and Eg, and 
the angle included by them becoming known, the 
angle EOg — OgE may be found from the proportion 

EO + Eg :EO - Eg : : tan i{EOg + OgE) : tan 

i(EOg - OgE), - . 
and finally 

EOg = i{EOg + OgE) + \{EOg - OgE). 

solutions of examples. 
Page 169. 

1. P will move down. Solving for s in Equation (3), 
p. 166 of text,, we have 

__ P sin B - TFsin A % 
S ~ 2(P + W) gt 

5347-71 AQaf , 
= — YJ?) — - 4:8-6 feet. 



128 KEY AND SUPPLEMENT 

2. If the weights be in equilibrium, v = and 

Eq. (1), p. 105, becomes 

A PsmB - Ws'mA a 

° = P^rw %» 

or 

PsmB= TTsin^. 

3. In this example the acceleration = \g ; 

which in Eq. (3), p. 166 of the text, gives 
1= itP+ W) 



P sin 45° - TFsin30 3 ' 
or 

P. x 0-7071 -iW= i(P + W) ; 

... p = i.38 ir+. 

4. From the conditions of the problem, and the 
figure, we have 

CA = 3 ft., AE= 1 ft, FB = J ft. ; 

.-.OF =2 ft, CF=2ih.,EF=Uh. 

The cylinders being of the same material, 
their weights will be as the squares of their 
radii ; hence 

Wt = 4 % 

Take the origin of moments at g ; or since 
gC is vertical, the moment will be the same if 
the origin be anywhere on that line. The 
arm of W\ will be the perpendicular from E 
to the line Cg, which is 



TO ELEMENTARY MECHANICS. 129 

Eg . sin EgC = Eg . sin cp. 

Similarly, the arm of W 2 will be 

gF. sin g>. 

Hence the equation of moments will be 

Wi . Eg sin cp = W 2 . ^i^sin <p, 

or, substituting the value of Wi and cancelling 
sin ^, 

±W 2 .Eg = W 2 .gF, 
.: \Eg = gF. 
But 

Eg + gF= l^ft.; 

substituting, 

5^ = 1|- ft. ; 

s. Eg = & ft 
and 

9^=1 ft. 

From trigonometry we have 

cos CEF - (CZ?+{EF?-(CF)* 

C0S L ^ ~ WE.EF 

_ i + 2j- - 6} 

- g -o, 

.-. CSF= 90°, 
and "\ve have 

tan i^ = -^=4 = 0-15; 
.*. ECg = S° 31' 50" +. 
5. The mass moved remains the same, but the 



130 KEY AND SUPPLEMENT 

effective moving force is reduced by the 
amount of the friction ; hence we have at 
once 

JVP — W sin A- /*JFcos^ "1 



— 




p 


+ W 






-^ 


p 


— 


lF(sin 


A + \x cos 


ij) 


^. 




.]y 


P 


+ TF 






._„/ 


T 


j 


P + 


TF 







IF(sin A i- m cos ^f) </ J 

6. Let 72 = ^(7, r = AE = ,F5 ; 
.-. EC=P - r = CF, EF = 2r. 
EF is bisected by a perpendicular from (7; 

.-. sin CEF = -^ — . 

si — V 

By moments, as in Example 4, we find 

E (/ = iFF=%r. 

Since Cg is not perpendicular to EF, the tri- 
angle is oblique, and we have 

EC + Eg: EC- Eg : : tan ^Cy + ^C): 

tim&ECg -EgC), 
or 

i2 - fr : i2 - fr : : tan J(180 - E) : tan \{ECg 
- EgC) ; 



TO ELEMENTARY MECHANICS. 131 

from which the angle ECg may be found. The 
angle may be found by means of right-angled 
triangles. Thus, let fall a perpendicular from 
C upon EF, and let t\\a foot be represented 
by G, which the reader can supply in the fig- 
ure. Then 



CG = V(£ - rf - r\ 
and 

tan ECG = 



.-. ECG = tan- 1 / r =- 
Also 
tan gCG — —^ 



V(B-r) 2 - r* ' 
... g CG = tan- 1 - 



3 Vi2 2 - %Rr 
and, finally, 

ECg - ^C# - gr<7^. 

7. The velocity of discharge will be 

v = Vtyh, 
and the quantity discharged in one second will 
be 

k V2gh, 
and in 'the time t the quantity of discharge 
will be 



232 KEY AND SUPPLEMENT 

Id Vtyk] 
hence 

solutions of examples. 

Page 179. 

1. The formula, p. 171 of text, becomes 



50 



cos cp 

sin 30° 

cos (60° - 30°) 

= 28-8 lbs. 
2. We will have 

cos (—45 ) 



3. We find 



F = W ^A- TF sin ^4 lbs. 
cosO 



4. We will have 



_sin^_ = W tan ^ lbs. 
* cos(-^) 



TO ELEMENTARY MECHANICS. I33 

5. From Article 254 we have, in reference to mo- 

tion down the plane, 

F= ^-f- 2 n (3 °? ^ 60 = 18-1 lb,.; 

cos (— 5 ) — 0-2 sin (— 5 ) 

up the plane 

F= ^ O - 2 ; o ; 30 ; o 50 = 344 lbs. 

cos 5 —0-2 sm 5 

Page 180. 

6. We will have, from the first value of F, p. 172 

of the text, 

F = i ya - 0.15 x iV 2 1Q0 = 6Q ]bg _ 

T. From Article 256 we have 
sin A = Jf- 

But from Example 4, p. 146 of the text, 
where r — \R, we find (see solution in this 
Key) 

.*. sin A = ^ ; 

.'.A=r 35' 40. 

8. From the formula of Article 259 we have 

Cc = if x 4 = a feet, 

and, therefore, the three sides of the triangle 
OCc become known ; hence, from trigonome- 
try, we have 



134 KEY AND SUPPLEMENT 



ab 
~ r 12 

.-. 0=28° 51' 20." 

9. The formula of Article 259 gives 

P = w. 

The triangle will be equilateral, hence each of 
the angles will be 60°. 

10. The equation on page 179 gives 

F=CD = == 

2V1-A 

= y'S feet. 

answers to exercises. 

Page 180. 

1. It will. 

2. It cannot without a force to hold it. 

3. When the centre of gravity is highest the poten- 

tial energy will be greatest, and least when 
lowest. In Fig. 113 it is greatest. 

4. In Fig. 114 the potential energy is a maximum. 

In Fig. 115 the potential energy is least. 

5. In Fig. 117 it is indifferent, for, since the weights 

are in equilibrium at all points on the curve, 



TO ELEMENTARY MECHANICS. 135 

if the weight W be moved from one position 
to another on the curve, the weight P will be 
raised or lowered so that their common centre 
of gravity will remain at the same height. 

The same general relations hold in Figs. 118 
and 119 ; hence the potential energy is indif- 
ferent in these also. 

Page 181. 

6. In this case they are in equilibrium in only one 

position. If P > W, we will have e > 1, and 
the value of y will be imaginary ; hence W 
necessarily exceeds P. 

7. They cannot, for the eccentricity would be less 

than unity. 

8. If the weight IF be at the upper extremity of 

the axis, at A, there will be equilibrium. 

9. The length will equal twice the diameter of tli£ 

bowl, and it w\\l rest on the edge F. 

10. Horizontal. 

11. It would rest on the horizontal plane. 

Page 186. 

The expression for the moment in the 4th line of 
page 186 should be 

F\{x\ cos A — y 1 cos a j), 
for reasons given in Article 176 of this Key. This is 
the same as 

F t (a?! sin a x — y x cos a^). 

The signs of all the other expressions on that page 
should also be changed from + to — and — to +. 



136 KEY AND SUPPLEMENT 

solutions of examples. 
Page 192. 

1. t = 9 ]'*« 100 = 90-5 lbs. 

2 x 0-bbS3 

2. Let AB = x, BC = y, 

AD = a, DC=v, 

t = the tension of ^4i?, 
£j = the tension of BC; 
then 

» + y = io, (l) 

u + v = 5, (2) 

* == 24 ; (3) 

and the last formula of Article 269 of the 
text gives 
sec BAD = 2 sec j9CZ>, 

or 

- = a£. (4) 

From the figure, 

a? - ^ 2 = ?/ 2 - f 2 = j?Z> 2 . (5) 

Eliminating between equations (1), (2), and 
(4) gives 

Equations (1), (2), and (5) give 



TO ELEMENTARY MECHANICS. 137 

T> - u 2 = (10 - xf - (5 - uf, 
or 

4tx -2u= 15. (7) 

Combining (6) and (7) gives 

x = 4-47 + ft., 
t* -144 + ft.; 
.-. y — 5-53 + ft., 
v = 3-56 + ft. 

Page 193. 

3. These conditions require that the points A and 

C shall not be in the same horizontal ; and in 
the values of t and t t , page 188 of the text, 
BAD — 0, and we have 

t= W cot BCD, 
ti= W cosec BCD 
W 
~ sin BCD ' 

4. The angle DAC will be 45°, and the last formu- 

las on page 191 of the text give 

c cos 45° = t 
c sin 45° = 250 lbs. ; 
250 



' 0-7071 



= 353-5. lbs. 



250 n __ 

t = mn x0 - 7071 

= 250 lbs. 



138 KEY AND SUPPLEMENT 

5. The last equation on page 191 of the text gives 

WsinCA&^iW; 

.--. sin CAD = £ ; 
hence 

CAD = 30°, 

and the depth CD will be one-half the length 

of the rafter. 

6. The equation 

t = iW cot CAD, 
page 191 of the text, gives 

W=iWcoiCAZ>; 

.-. cot CAD = 2 ; 

.-. CAD = 26° 33' 54". 

answers to exercises. 

Page 193. 

1. They will not. The two added forces R and — R 

will form a couple. 

2. They will not, for the resultant of F and F x 

combined with — R will constitute a couple. 
If the force equal to R were to act in a direc- 
tion opposite to the resultant, then, in Exercise 
1, there will be a resultant equal and oppo- 
site to the fourth force, and in the 2d Exercise, 
after the 3d force be removed, there will be a 
resultant equal to 2R. 

3. No modification is needed, for the moment of 



TO ELEMENTARY MECHANICS. 139 

the components will equal the moment of the 
single force. 

4. It cannot — to do so would require an infinite 

tension. 

5. If the weight at B is free to adjust itself, the 

tension on each will be equal, and each equal 
tofTT. 

6. The tension will remain the same. The tension 

is dependent only upon the weight and slope 
of the parts, as shown by the equations on 
p. 188 of the text. 

7. An ellipse for the sum of the distances from the 

fixed points A and C will constantly equal the 
length of the string. 

8. Decreased — and the thrust at the lower ends 

will be diminished. See the last equations of 
Art. 271 in the text. 

9. The thrust and stresses on the braces will both 

be increased. 

10. If the strut supports the weight, there will be 
no stress on the rafters. 

Page 197. 

Galileo was the first writer, of whom we have any 
knowledge, who established formulas for the strength 
of beams. His work was published at Bologna in 
1656. Although the hypotheses upon which the for- 
mulas were founded were false, yet the law of variation 
of strength which he deduced for rectangular beams 
was correct. This law is — the strength varies directly 



140 KEY AND SUPPLEMENT 

as the first power of the breadth and the square of 
the depth jointly, and inversely as the first power of 
the length of the beam. But the factor used by him 
for determining the value of the strength was three 
times too large. 



solutions of examples, 

Page 201. 

1. Let io = the load per unit of length of the 

beam, I = the length of the beam, then will 

W= wl 

We have from the last equation on page 200 
of the text, 

= 3 500 x 8 x S x 12 
" ¥ 1400 x 8 x 8 
= Sj 3 ^ inches. 

Remark. — It is best to reduce all the dimensions 
to inches, since the tabular value of JR, is given 
per square inch. 

2. From equation (2) of Art. 2S2, we have 

7 72 GPl 

But 

d = 4b ; 



TO ELEMENTARY MECHANICS. 1^1 

PI 

_. ;l 1000 x 8 x 12 
~ ' 8 1200 

= 30; 
.-. lj = 3-10 + inches ; 
and 

(I = 12-43 + inches. 

3. From Problem 3, p. 200, we have 

d ~ 2 M 

_3 8000 x 12 x 12 
~/\ ' 2 x 10000 
= 864; 
.-. d = 9-29 inches. 

4. From the same equation as the preceding, we 

have 

M ~- 2 bcP 

_ 3 20000 x 10 x 12 
_¥ 4x9x9 

= 11111-1 + lbs. 

5. The required stress will be the value of R found 

from the equation above, 



142 KEY AND SUPPLEMENT 

. 7 _ 2 Eb& 

_ 2 20000 x 1| x (3|) 2 
~ 3 " 10000 

= 24-5 inches. 

6. Problem 4, p. 200 of the text, gives 
Bbd* 



A 12000 x 6 x 144 
~ *~ 15 x 12 

= 76,800 lbs. 

7. The load will be uniform, and will equal the 
weight of the beam. We have 





W = 


2 x 2 x 


I x i 












-.1, 










and the formula of 


problem 


2, 


P. 


199 of the 


text, 


gives 
1 = 

.-.? = 


fibd? 
-3 w 

1 30000 

- 3 

: 80,000; 


x 2 x 4 
I 


? 






and 
















■Z = 


= 282-8 inches 









= 23 feet 6-8 inches. 



TO ELEMENTARY MECHANICS. 143 

Page 205. 

The straight line of quickest descent is not the line 
of quickest descent. Curves of quickest descent are 
called Brachistochrones. Their form depends upon 
the conditions assumed. The forces may be assumed 
to vary according to the inverse squares, or directly 
as the distance, or inversely as the distance, or ac- 
cording to some other law, and they may be assumed 
to act in parallel lines or radiate from a point. If 
they are constant and parallel, as in the case of terres- 
trial gravitation, the curve will be a cycloid. It was 
problems of this character that gave rise to the Cal- 
culus of Variations — a very high order of analysis. 

solutions of examples. 
Page 208. 

1. From the 3d equation, p. 203 of the text, we 

have 

9* 
200 

32i x 25 

= •2487; 
.-. cp =: 14° 24'. 

Page 209. 

2. From the 2d equation on p. 204 of the text we 
' have 

s = v Q t — \g$ sin cp ; 
... s = 50t - 16^- xi^t 2 

= 50£-'8 s >W 2 ^ 



sin cp -. 



144 KEY AND SUPPLEMENT 

Hence, in 3 seconds 

s = 150 - 72|a/2 = 47-65 feet. 

At the end of 5 seconds 

s = 250 - 201 A- V2 = - 34-3 ft., 

that is, it will have ceased to ascend the plane, 

and returning, will, at the end of 5 seconds, 

be 34-3 feet below the starting-point. 

At the end of 10 seconds 

s =1 500 - 804JV2 = - 637-3 feet below the 

starting-point. 

3. The required velocity will be the same as that 
acquired by the body in sliding down half the 
length of the plane ; hence the required veloc- 
ity will be 

v — Vgs sin <p 



= a/321 x loo x T 2 ¥ °o 
= 25-36 feet. 

If a body starts from the middle of the plane 
at the same time as the one at the upper end, 
it will reach the foot in the same time that the 
upper one reaches the middle; hence, if it be 
projected upward with the velocity acquired, 
at the instant the body starts from the upper 
end, they will meet at the middle of the plane. 
4. The point must be higher than the lower ex- 
tremity of the diameter — otherwise the solu- 
tion is not possible. The required line will be 



TO ELEMENTARY MECHANICS. 145 

the distance from the point to where the line 
cuts the circle when drawn to the extremity 
of the diameter. 

5. From equation (2), p. 208 of the text, 

= i V22-4 x 5280 

= 38-2 -f feet per second 

= 26-05 + miles per hour. 

(To reduce feet per second to miles per hour, 
multiply the former by *£^? = J|.) 

G. From the 5th equation we have 

and from the 2d (writing s' for s so as to dis- 
tinguish it from the preceding s) 

v* = &(h - 17-6)*' 
= h x 324 x 2640; 
4-6 x 82-4 x 2640 

''' s = a 

= 4857-6 feet. 

7. The time down the plane may be deduced from 
the 3d equation of the text. We have 



t : 



^/~328^ 



h - 17-6 

4 / 328 x 2640 
" r 32-4 



146 KEY AND SUPPLEMENT , 

Page 209. 

= 163-4 sec. 

= 2-72 minutes. 

For this time on the horizontal, we have from 
the 4th and 5th equations 



< = ^± 



= 90/^ 

3 r a. a 



/4858 
4-6 

= 303-3 seconds, 
= 5-05 + minutes. 



8. Equation (5) will give the velocity which it 
must acquire in moving down the plane. We 
have 

2_ 8 

= 1000 
4-6 
= 217-3 feet per second. 

The required height will be given by equation 
(2) ; we have 

s 
_ 81 x 217-3 + 17-6 x 1200 



1200 
= 32-27 feet. 



TO ELEMEXTAKY MECHANICS. \^ 

9. Equation (5), page 208 of the text, gives 

v = V — 
r 4-6 

y 4-6 
= 13-19 ft. per sec. 

Equation (2) gives 

81^ 



h - 17-6 

_ 81 x 173-91 
7-4 

= 1903-64 ft. 

The author once had occasion to use the prin- 
ciples of the last example in constructing the 
approach to an ore dock at Marquette, Mich * 

solutions of examples. 

Page 217. 

According to Article 299, the range will be 

2 A sin 2a, 
where 

hence the range will be 

— siniJa 



-^g KEY AND SUPPLEMENT 

= ?^! sin 90° 
9 

= 8041 feet. 

The greatest height will be, (Art. 301) 

h sin 2 a 

= *- sin 2 45° 

= " 6J- x 32i 

= 201-04 + feet. 

Page 218 

2. 3C=%</h.y, 



= 2a/15 x 12 
= 26-8 + feet. 

3. From Article 299, we have 

sin 2<* (/ 
which in Article 300 gives 

tf = — sin 2 a, 

A(j sin a cos <* 

or solving for tan a gives 

gt 2 
tan ar = £ — 



TO ELEMENTARY MECHANICS. 149 

3 2|- x 225 
~ 2 x 1000 

= 3-6187 

.-. a = 74° 33' 9''. 



Page 218. 

From Article 300, we have 

v— ?r 4 — 
A sin a 

32|- x 15 
~2 x -96387 

= 250-29 feet. 

From Article 301, we have 

CD — It sin 2 a 

_(250-29)% nc 



B 


- ^ yy „y„„ t j 




= 901-69 feet. 


:. From Article 299, we have 




h - X - V% • 




2 sin 2a Zg' 




i/2 x 32i x 25000 

* * V ~ y 2x1 




= 896-8 + feet. 


From 


Art. 301, we have 




A = h sin 2 a 



150 KEY AND SUPPLEMENT 

= 6,250 feet. 

From Article 300, we have 

_ 2# sin «r 
9 

_ 2 x 897 x W2 

= 89 + seconds. 

5. We have from Articles 299 and 301, 

AB = WD, 

or 

4A sin a: cos a = 4h sin 2 <*, 
dividing by sin a cos a, 
tan « = 1 ; 
.-. a = 45°. 

6. In equation (3) page 214, make y = — 150, 

which is negative, because in Fig. 139 y is 
positive upwards, and the point where it will 
strike the plane, in this example, is below the 
point of starting, tan a = 1, and we have 

- 150 = x - *f* , , 

2(7o) 2 x i ' 
which solved for x gives 
x= 271-5 feet. 



TO ELEMENTARY MECHANICS. 151 

7. In this example 

a= - 30°, y = - 25 feet. 
The velocity with which the body will leave 
the eaves will equal that of a body falling 
through the vertical height of the ridge above 
the eaves, and this value will be considered as 
the velocity of projection. "We then have 

v> = fyx7=Ug y 

and these substituted in equation (3), page 214, 
give 

-25= -0.57735*- ^ x ^ 6Q3 , 

which solved gives 

# = 17-6 + feet. 

8. Substituting the values given in the example 

for a and y in equation (3) page 214 gives 

60 = 400 tan, -fj^f; 

50 = 600ta n «- 3 jy 60 f . 
2tr cos 4 a 

Multiplying the first by 9 and the second by 
4, and subtracting the latter from the former, 
gives 

340 = 1200 tan a; 
.-. tan oc= £J, 
and 



152 .KEY AND SUPPLEMENT 

a = 15° 49' 9". 

Substituting this value in the first of the pre- 
ceding equations gives 

60 - 11 x 400 32i(400£__. 



'2 x (0-96213) 2 x tf : 






x 400 



320(0'96213) 2 
= 228-2 feet per second. 

exercises. 

Page 219. 

1. Zero. 

2. The velocity of projection being the same, they 

will strike the sea at the same time, and their 
range from the point where the ship will be at 
that time will be the same ; but not the same 
if reckoned from the point of projection. 

3. 15 miles per hour = 1 fjxtf- feet per second 

= 22 feet per second ; hence the actual veloc- 
ity will be 11 feet per second in the direction 
of motion of the ship in reference to the point 
from which the projection is made. 

4. In reference to the point on the earth, it will be 
t the same ; but not in reference to the point in 

space from which the projection is made. 

5. It will reach it in the same time. A horizontal 

motion does not affect the time of descent due 
to gravity. The projectile falls from the 



TO ELEMENTARY MECHANICS. 153 

highest point of its path (in a vaeunm) in the 
same time that it would fall vertical down- 
wards. 

6. They will ; for according to Article 304 of the 

text, we have for equal ranges the angle a and 
90° — a. Let a — 45° — 6 ; then will the an- 
gles be 45° - S and 90 - (45° - 6) = 45° 
+ £; but 45° -f S is the complement of 45° 
- &. 

7. The lines will be the sides of an angle, and since 

the velocities are uniform, they will be divided 
into equal parts in equal times, by the motion 
of the bodies; hence, by geometry, the lines 
passing these equal divisions will be parallel. 
Page 223. 

The relation between centripetal and centrifugal 
forces has been the subject of much discussion. In 
an article which appeared some time since in "Na- 
ture" it was asserted that the term centrifugal force 
had done much harm in mechanical science, and 
ought not to be used. The basis of the trouble with 
such writers is, they consider that centrifugal force is 
to be applied to the same body as the centripetal ; but, 
as stated in the text, such is not the case. Centripe- 
tal force is generally conceived to be the action of the 
ruling, or larger body, upon the smaller one, while 
centrifugal force is the equal opposite action upon the 
other body. Thus, if a nail holds one end of a string, 
while a body attached to the other end is made to ro- 
tate rapidly about it, the nail represents the ruling 
body, and the centripetal force is the pull of the string 
upon the rotating body, and the centrifugal force the 



154 KEY AND SUPPLEMENT 

pull of the string upon the nail. Similarly the at- 
tractive force between the eartfi and sun is centripe- 
tal if applied to the earth, and centrifugal if applied 
to the sun. These are illustrations of the third Law 
of Motion — that action and reaction are equal but 
1 opposite. 

A constant centripetal force, caused by uniform 
motion in a circle, does not produce an acceleration, 
for it does not act along the same line. The action 
being constantly normal to the path, its effect is con- 
stantly expended in deflecting the body from a recti- 
lineal path. Should it cease to act as soon as the 
deflection is made; the body would move in a right 
line, in accordance with the First Law of Motion. 

The analysis for determining the value of the cen- 
trifugal force in the text is lengthy but strictly logical. 
The following solution may be more acceptable. As- 
sume that the body is moving around 
the circle at a uniform rate ; then 
will the centrifugal force be con- 
I stant, and at any point, as B, the 
direction of motion will be that of 
the tangent BD. The centripe- 
tal force must be such as to draw 
the body from the tangent BD a 
distance equal to DC in the same time that it would 
move over BD. Draw EC and CG, and a chord BC. 
If now the point C be indefinitely near B, the limit 
of the sine EC (or its equal BD), and of the chord 
BC will be the arc BC. 

Let BD be the space moved over in time t, and v 
the constant velocity, then 




TO ELEMENTARY MECHANICS. 155 

V 

•since the motion in the arc is uniform ; and since the 
centrifugal force is constant, the space BE will be 
given by equation (2), p. 12 of the text, or 

BE = }v% 

where v' is the velocity which would be produced by 
the centripetal force in passing over the space BE in 
time t, if the force acted along the line BE. But 
since the times are equal, we eliminate t, between 
these equations, and find 

, 2BE 

v =~ec v ' 

From the figure we have, since EC is a mean pro- 
portional between BE and EG, 

(EC) 2 = BE(BG - BE), 

which ultimately becomes 

(EC) 2 = BE.BG; 

.-.BE-- 



(EC) 2 



BG ' 

Substituting this value above gives 
, 2EC 

* = !&» 
EC 



156 KEY AND SUPPLEMENT 

since BG = 2/\ But ultimately the velocity along 
i?2) or EC is that along the arc BC, and ultimately 

^(7 = vt, 

where n is tlie velocity along the arc. This substitu- 
ted, gives 

*2 



V _ Tf 

1 ~~~r 



or, multiplying by m, 

mv' V 2 

But the left member is, according to Article 122, the 
value of a constant force, hence 

the required result. 
Page 229, Articles 319, 320. — Sir Isaac Newton con- 
ceived the fact that if the attraction of gravitation 
varied as the inverse square of the distance from 
the centre of the force, it ought to account for the 
motion of the moon ; that is, the force of gravity ex- 
erted by the earth should just equal that necessary to 
cause the proper deviation of the moon from a tan- 
gent to its orbit. His first efforts to prove this law 
tailed, due to the fact that an erroneous value of 7?, 
the radius of the earth, was used. Instead, however, 
of abandoning the idea, and attempting to account 
for the motion according to any other hypothesis, he 



TO ELEMENTARY MECHANICS. 157 

returned to his calculation from time to time, but 
with no better results. Finally, while attending a lec- 
ture in London, he obtained a corrected value of the 
radius, which, when substituted in the equation he 
had so often reviewed, established his theory. He 
was so overcome by the grandeur of the problem as 
the final proof was becoming apparent, that he was 
unable to complete the numerical reduction, and 
called a friend to do it for him. 

It will be seen, in Article 319, that the radius of 
the earth enters the formula, in determining the dis- 
tance of the moon, in the expression 60-367?. The 
value of R which he at first used was too small by 
tV t° tt °^ ^ s true value. See also remarks on pp. 
36 and 37 of this Key. 

The law of gravitation was not, at once, universally 
accepted. Several times, especially in the history of 
astronomy, certain phenomena appeared to conflict 
with this law, when it was called in question, and its 
truth assailed. But all opposition to it disappeared 
after Laplace, by his truly wonderful analysis, ex- 
plained all those paradoxes, and accounted for all the 
motions of the solar system, on the simple law of 
Universal Gravitation. It is now believed to be true, 
not only for the solar system, but for every particle 
of matter in the universe. Newton believed that the 
ether of space, whatever it might be, was more dense 
in the vicinity of the planets, than in remote space ; 
that, indeed, it might be only air extremely rarefied. 

To find the stress due to the attraction "between the 
earth and moon. 

It equals the centrifugal force, the value of which is 



158 KEY AND SUPPLEMENT 

moo 2 r. 

The mass m of the moon is about 3^ times that of a 
mass of water of equal volume, and as a cubic foot of 
water weighs 62£ lbs. when g = 32£ feet, and the 
diameter of the moon is 2,160 miles, we have 

m - 3i X ^(2160 x 5280) 3 X ?* lbs. 

The time of the revolution of the moon about the 
earth is about 27^ days ; hence the angular velocity 
per second is 

- 2tt 

00 ~ 27i X 24 x 3600 ' 

The mean distance between the centres of the moon 
and earth is about 240,000 miles ; hence 

r = 240,000 X 5280 ; 

all the magnitudes being in feet, and all the times 
reduced to seconds. Hence we have 

_ 3j x J7T 3 x (2160) 3 X (5280) 4 X 240000 x 62 j 
stress - 82 ^ x (2 ^ x 24 x 360Q , 2 

which reduced gives, approximately, 

44,000,000,000,000,000,000 lbs. 
= 44 x 10 18 . 

A steel rod one square inch of section will sustain a 
pull of 120,000 lbs. ; hence it would require (approx- 
imately) 



TO ELEMENTAKY MECHANICS. 159 

370,000,000,000,000 
= 37 X 10 13 square inches 

of steel to bold the moon in her orbit if substituted 
for the attraction between the earth and moon. 
In one square mile are 

4,014,489,600 sq. 'inches; 

which, divided into the above, gives, for the equiva- 
lent section in miles, 

90,000 sq. miles nearly. 

Since the radius of the moon is 1,080 miles, the area 
of a great circle will be 

3,660,000 sq. miles nearly, 

which, divided by the solid section of the steel rod, 
gives 40 -f- ; hence, if the rods were each one square 
inch in section, and the great circle of the moon be 
divided into inch-square spaces, the rods would cover 
one space in 40. 

The square of the diameter of the earth is nearly 
15 times the square of the diameter of the moon, hence 
such a steel rod would cover about g-J-g- of the merid- 
ian circle of the earth. 

If the material be iron instead of steel, and if 
10,000 lbs. be taken to represent the tenacity, a value 
quite commonly used 'in engineering structures, the 
rod — or rods — would cover more than one-fourth the 
cross-section of the moon, and about -fa of a great cir- 
cle of the earth. 

The same problem applied to the attraction between 
the sun and earth gives 



Ig3 KEY AND SUPPLEMENT 



5^ x -J7r 3 (20500000) 3 X 5280 X 92 5 00000 X 62j ,. 

321 x (365 X 24 X 3600) 2 ' 



where it is assumed that the mean density of the 
earth is 5-j- times that of water, the distance between 
the centre of the sun and the earth 92,500,000 miles, the 
radius of the earth 20,500,000 feet, and the time of 
the revolution in the orbit 365 days. This reduced 
gives, approximately, 

912 x 10 13 lbs., 



912,000,000,000,000,000,000 lbs., 

or more than 20 times that between the earth and 
moon. According to this result it would require a 
solid steel rod of a cross-section equal nearly to one- 
half the great circle of the moon, the tenacity being 
120,000 lbs. per square inch ; or if the rod be of iron, 
and 10,000 lbs. be used for its tenacity, the section of 
the rod will be about f of the area of a great circle of 
the earth. These examples show the immense stress 
of gravitation when large masses are involved. 

The following examples will show that the same 
force, under certain circumstances, is comparatively 
weak. 

Required the time that it will take two spheres of 
the scone material as the earth, each one foot in diam- 
eter, placed 12}- inches from centre to centre, to come 
together by their mutual attractions, in void space. 



TO ELEMENTARY MECHANICS. 161 

According to one of Newton's laws of attraction, 
the force varies as the mass. If the diameter of the 
earth be 41,700,000 feet (p. 33 of the text), then will 
the mass of the sphere 1 foot in diameter be 

1 



(41700000)* 



of the earth, 



and therefore the acceleration at the surface of the 
earth due to the attraction of such a sphere placed at 
the centre of the earth would be 

<7 



(41700000/ 



an inappreciable quantity. 

According to another of Newton's laws, combined 
with the analysis on pp. 33 and 34 of this Key, the 
force varies inversely as the square of the distance 
from the centre of the sphere ; hence at the distance 
of one foot from the centre of the small sphere we 
have 

1»: (20850000)' : = 8(20 J 0000) , :/; 



■v/ = 



9 

8 x 20850000 ' 



for tne acceleration of a particle one foot from the 
centre of the sphere. This will also be the accelera- 
tion of any uniform sphere whose centre is one foot 
from the first sphere, if the diameter of the second 
sphere does not exceed one foot ; or should it exceed 
that diameter, it will still be true for the mass of the 
sphere if reduced to a sphere whose diameter is less 



162 KEY AND SUPPLEMENT 

Page 229. 

than one foot, and hence will be the acceleration pro- 
duced upon an equal sphere. If the first sphere were 
fixed in space, the second sphere would move the J of 
an inch ; but as both are free each will move one- 
half the distance between them, or £ of an inch. 

In order to simplify the problem, we will assume 
that the acceleration is uniform, while the spheres are 
moving the \ of an inch, and is that due to the at- 
traction at a distance of 12 inches from their centres ; 
then will equation (i), page 12 of the text, be applica- 
ble, and we have 



Y% 



V: 



2 x i x T y 



8 x 20850000 

= V108031 

= 328-7 seconds, nearly, 

or less than 5^ minutes. 

This problem is in " The System of the World," by 
Sir Isaac Newton, p. 527 of our copy of the Principia. 
It is there stated that " the attraction of homogeneous 
spheres near their surfaces are (Prop, lxxii.) as their 
diameters. Whence a sphere of one foot in diameter, 
and of a like nature to the earth, would attract a 
small body placed near its surface with a force 
20,000.000 times less than the earth would do if 
placed near its surface, but so small a force could pro- 
duce no sensible effect. If two such spheres were 



TO ELEMENTARY MECHANICS. 163 

distant but J of an inch, they would not, even in 
spaces void of resistance come together by the force of 
their mutual attraction in less than a month's time." 

We have sought for the source of the error in the 
Principia, by determining the conditions necessary 
for giving his result, but have not satisfied ourselves. 
"We observe that he made an error in saying that the 
force of attraction on the surface of the small sphere 
is 20,000,000 times less than on the earth ; for, ac- 
cording to his proposition — considering the radius of 
the earth as 20,000,000 feet— it should be 40,000,000 
times less ; and according to the inverse squares, at 
the distance of one foot from the centre of the small 
sphere, it would be 160,000,000 times less. If now 
we assume that the particle is moved \ of afoot, un- 
der the action of this force, we would have 



>-y, 



2x 



32 



160000000 



= a/2,500,000 (nearly) seconds, 

where the quantity under the radical is nearly the num- 
ber of seconds in one month. "Whether this gives any 
clue to the source of the error, we are unable to say.* 

* The author presented the above result to the Physical Sec- 
tion of the Am. Asso. for Ad. Sc, at Montreal, 1882. The genu- 
ineness of " The System of the World, by Sir Isaac Newton," was 
there called in question.- The work was originally issued in a 
separate volume, but in the author's copy it is bound with the 
Principle/,, and paged with it. But it evidently forms no part of 
the Principia proper. 



104 KEY AND SUPPLEMENT 

Page 229. 

The assumptions made in order to simplify the 
problem, not being strictly accurate, we now propose 
the following problem : 

Assume that two equal spheres of the same material 
as the earth, each one foot in diameter ', are reduced 

in size to a mere point at their centres, and placed 
one foot from each other j required the time it would 
take for them to come together by their mutual attrac- 
tion, they being uninfluenced by any external force. 

We first make a general solution. 
Let E = the mass of the earth, 

m — the mass of one of the spheres, 
m — the mass of the other sphere, 
B = the radius of the earth, 
r — the radius of one of the spheres, 
t = the radius of the other sphere, 
a = the acceleration due to gravity on the 

earth, 
jj. = the acceleration due to the attraction of a 
sphere of mass unity upon another 
sphere ot mass unity, the distance be- 
tween their centres being unity, 
a — the original distance between the centres 

of m and m\ and 
x = the distance between their centres at the 
end of time t. 
The values of units are determined by measure- 
ments on the surface of the earth, and the unit mass 
will be so taken as to correspond with the units in 



TO ELEMENTARY MECHANICS. 165 

use. The acceleration produced by a mass E, con- 
ceived to be reduced to a point, upon one of the units 
of mass at a distance unity, will be E times as great 
as that produced by the other unit, or 

and at a distance H it will be 
E_ 

which, according to the notation, will be the accelera- 
tion on the surface of the earth due to gravity; hence 

g = v- jp . (i) 



and 



f = §ff, (2) 



by means of which the numerical value of the unit of 
acceleration may be determined. 

Again, the acceleration produced by the attraction 
of the mass m upon one of the units of mass at dis- 
tance, unity will be 

and at a distance, x, the acceleration will be 

M_m > (3) 



166 KEY AND SUPPLEMENT 

which will also be the acceleration produced upon m, 
by the attraction between m and m! at the distance x 
between them, since the result will be the same as if 
both masses were concentrated at their centres of 
gravity ; for all of m will exert the same force upon 
each particle of m' as upon each particle of the unit. 
But the pull in 2 :>01in ^ s w ^l equal the mass into the 
acceleration (p. 44, Art. 86 of the text), or 

mm' ... 

./"3T- W 

Similarly, considering the attraction of m' upon ra, 
the acceleration produced upon m will be 



*£ , (5) 



and the pull in pounds will be ?n times this amount, 
or 

m'm /ax 

M-jT> (6) 

which is the same as (4), as it should be. Substitut- 
ing jj. from (2) in (4) or (G) gives for the pull in 
pounds (or their equivalent), of any two masses m and 
m', 

mm R 2 q 1 ,_. 

— *-•?■ ( '> 

The origin of the axis of x being at the centre of 
one of the masses and moving with it, and the total 
mass moved by the stress being m + m', we have 



TO ELEMENTARY MECHANICS. 167 

, ,,rPx m?n'Ji 2 g 1 , . 

(OT+W )_ = ___^._, ( 8 ) 

which integrated (Analyt. MecJi., pp. 33, 34), observ- 
ing that for t = 0, x = 0, and v = 0, and that /* in 

the reference equals - — ■ —^ in this case, ffives 

<=[ fi l^'] t 4<--^ + "»-'© i ]/»> 

which for the limits gives 

f(m -f m')Ua~\i 
' = **«[_ ZmmUPg J * 

If both spheres are of the same density, their masses 
will be as the cubes of their radii ; or 



IP 



»'=Jr*;J 



^ (10) 



and Ave have 



and if the spheres are equal, as in the problem, we 
have 

If a = 1 foot, i? = 20,850,000, r = £ foot, g = 32J, 
we have 



168 KEY AND SUPPLEMENT 

- K7mQ /166800000\1 
==1 ' 57079 x . 32466 ) 

= 3577 seconds, nearly 

_ = 59-6 minutes, nearly. 

Page 229. ' * 

An exact solution of the former problem may be 

made by means of equation (9), by substituting in it 

a = 12t inches = 1-h feet, and making x — a for one 

limit and 1 foot for the other. We would thus have 

. (Ik x 20850QQ0\i r ' , / ISM n ~| 

'=( rrrooT- (1^-1) + 1ACOS-1.- -0 



= 384 seconds nearly, 
or less than 6J minutes. 

The following is the reduction of the preceding ex- 

/ 12 \i 

pression. To find the value of cos - M ^- J 5 we have 

log. 12 =-1.079181 
log. 12-25 = 1-088136 

Dif. = 1-991045 

Dividing by 2, 1-995523, 

and log. 0-9S974 = f-995522, 

or log. cos 8° 12' 46" = 9-995523. 

The length of arc will be 

8° 12' 46" - lg 

180° X 60 X 60 



TO ELEMENTARY MECHANICS. 169 

29566 Q1/11A . 
648000 8 ' U16; 

which may be reduced as follows : 

log. 29566 = 4470892 

log. 3-1416 = 0-497150 

ar. co. loff. 648000 = 4-188425 



subtracting 10, log. 0-14337 = 1-156467 
adding log. 1&, log. 1-02083 = 0-008951 



gives log. 0-11636 = 1-165418 ; 

hence, 

U- cos- 1 (j^-V = 0-14636. 

We also have (lh - 1) 1 = -h = 0-02083 + which 
added to the preceding result gives 0-16719 + for 
the value in the brackets. 

For the first parenthesis, we have 

log. lh = 0-008951 

log. 20850000 = 7-319106 

log. 8 = 0-903090 

log. 32i ar. co. = 8-492594 



subtracting 10 and dividing by 2, 6-723741 



log. 2300-7 = 3-361870. 
Adding log. 0-16719 = 1-223209 



gives log. 384-6 = 2-585079 ; 



170 KEY AND SUPPLEMENT 

that is. the time will be 385 seconds nearly, or less 
than 6^ minutes. 

To find the stress in pounds which would be ex-, 
erted by the mutual action of two such spheres at a 
distance of one foot between their centres, we have 
from equations (7) and (10), since m = m' 9 and x = 1, 

E 

By 

The mass of the earth is 5-j- times an equal mass of 
water. The weight of a cubic foot of water is 62£ lbs. 
at the place where g = 32-J-, and the volume of the 
earth is f nlP ; hence 

e = 5* x m x \7tip, 

which substituted above gives for the stress 

f n x 5| x 62| ,, 
20850000 x 321 bs '' 
= 0-00000215 + lbs., 

or less than TTrofo o o it °*' a pound, a quantity inappre- 
ciably small. 

A would-be inventor once proposed to weigh the 
varying force of gravity by means of very delicate 
scales, and by using them on board a steamer, thus 
determine whether the water underneath were deep 
or shallow. Since the density of the solid earth ex- 
ceeds that of water, the force of gravity at the sur- 
face of deep Avater will be less than on shallow water, 
but it is evident that the rocking and heaving of the 
vessel would probably produce more effect upon such 



TO ELEMENTARY MECHANICS. 171 

delicate mechanism than that due to the variations of 
the force of gravity. 

It is also stated that mariners have observed that 
two ships at rest in a quiet sea tend to approach each 
other, but it will be found that the gravitating stress 
due to their mutual attractions is so small that it 
might be more than neutralized by a very slight 
breeze, or by the beating of very small waves. 

To give some idea of the magnitude of this stress, 
assume that the vessels are of equal mass, and each 
equivalent to a sphere of the average mass of the 
earth, each 30 feet in diameter, and 200 feet between 
their centres. 

Conceive that the masses are reduced to their cen- 
tres, then since the mutual attraction of unit-spheres 
at distance unity between them is tottitowo °^ a pound, 
the attraction of the masses of the ships, reduced to 
their centres, at the same distance (one foot) will be 

imM m X 30 3 x 30 3 = 1471 pounds ; 

and at the distance of 200 feet, it will be 

1471 

_ r z= 0.04 pound nearly. 

If the distance between them be 500 feet, the stress 
would be 

5QQ2- = tto of a pound, nearly. 

If all external forces, such as the wind and action of 
the sea were neutralized, this slight stress would be 



172 KEY AND SUPPLEMENT 

sufficient to cause the ships in question to collide in 
a short time. 



solutions of examples. 

Page 233. 

1. We have from equation (5), page 226, 

tf TV v 2 
f=m- 







r 


g r 


Bi 


at/ 


must equal 


2TF; 






.\2TT= 


W v\ 
g r \ 






.'.v — 


V2^F; 



or the velocity must be that acquired by a 
body falling freely through a distance equal to 
the radius of the circle (see eq. (3), p. 36 of 
the text). 

2. If the tension is 3 TF, we have 

g r 

.w = VSgr. 

Let n be the number of revolutions, then the 
velocity will be the space (2?mi) divided by 
the time, or 60 seconds, r must be reduced to 
feet and the time to seconds, for g is given in 
feet per second. Hence we have 



TO ELEMENTARY MECHANICS. 173 



%nm nrn 
30~' 



V 


^"eo"^ 


nrn 
"30" 


= Vs^; 


»\n 


= 30/l 
n r 




30 
3-1416 A 




= 38-3. 



Vi6 T v 



3. The centrifugal force will equal the weight; 

hence 

v 2 W v 2 s* 
W = m - = — • - ; 
r g r 

,\ v = ygr = -^- (see preceding example) ; 

" 3-1416 y t 
= 38-3, 

as in the preceding example. 

4. When the body is at the lowest position its 

weight will be added to the centrifugal force, 



174 KEY AND SUPPLEMENT 

but the tension due to the centrifugal force 
equals the weight ; hence the tension will 
equal 2 W. 
5. The friction will be }x times the pressure due to 
the centrifugal force, and must equal the 
weight. Let n be the number of revolutions 
per minute ; then will the angular velocity per 
second be 

Inr . n __ 27tn m 
and the centrifugal force will be (Art. 314) 

and the friction will be 

f _ »nh?WR _ w . 
V- (30)V ~ W * 

30 / g 



The body is assumed to be in a radial groove, 
and the string slightly elastic so as to allow the 
body to move slightly along the groove, and 
thus give the friction a chance to act. 
The angular velocity per minute will be 

2tt x 250: 

and the centrifugal force will be 



TO ELEMENTARY MECHANICS. I75 

f _ TF/5007r\ 2 30 



g \ 60 J 12 

The friction will be 0-15 of this amount, and 
the tension of the string 0»85 of the same; 
hence the tension will be 

T= 0-85 x T Vs X - 6 -fM3.1416) 2 TF 
= 45-27 -f pounds. 

Paqe 234. 

7. The friction will be 

1 W" 
To VY ' 

The centrifugal force will be (Art. 313), 



hence 



Wi? 




9 J?' 




-git"™ 


w, 


-.9%. 

~;1G' 




_ i/m - 


- 2500 



10 
= 89-675 feet per second 
= 61 14 miles per hour. 



176 KEY AND SUPPLEMENT 

8. According to Article 322, we have 

7i-- - 
~ g ' r 

/ 30x5280 y 
V 60 X 60 / X 4a 
321 x 3000 

= 0-091 feet 
= 1-12 inches. 

9. The weight will be to the centrifugal force as 

the length of the string is to the lateral move- 
ment of the body ; or 

W:f:: G:%; 

.-,x= 6— 
W 

To find/we have equation (5), Art. 313, 



MO X 52S0\ 8 
_ W\ 60 X 60 ) 



g 4000 

which substituted above, gives 

_ 6 / 40 X 5280 N 

X ~ 321 x 4000 \ 3600 / 

= 0-160 ft. 



= 1-92 inches. 



TO ELEMENTARY MECHANICS. 



177 



The value is independent of the weight. 
10. To find the time of making one revolution, 
we have 

T= 60 4- 100= |. 

Then from p. 232 of the text, we have 



COS <p : 



4:7T 2 .AB 



~ 5 x (3-1416) 2 
= 0-23166 ; 
.-. cp = 76° 25' 13". 

Or by logarithms 

log. 32^ = 1.50T406 
log. (0.6) 2 =1.556303 

1 063709. 

log. 5 = 0.698970 
log. (3. 1416) 2 = 0.994300 



adding, 



adding, 1.693270 which subtract- 
ed from the above gives 

log. cos cp =1.370439; 

.-. <p = 76 = 25' 43 '. 

If cos cp = 1, or cp = 0, a relation will be es- 
tablished between AB and T, or we will have 



AB-- 



'4tt 2 ' 



and if A B is assumed to be less than the value 
8* 



178 KEY AND SUPPLEMENT 

found by this formula, cos <p will exceed unity, 
and hence cp will be imaginary, or, in other 
words, the conditions will be impossible. We 
also have 

AC = AB cos cp 

= 15 cos 76° 25' 43" 

= 15 X 0-234:66 

= 3*52 inches ; 
and 

BC = AB sin cp 

== 15 X sin 76° 25' 43" 

= 14-58 inches. 

11. Let n = the number of revolutions per minute, 
= n ~ 60 per second. The distance BC will be 

.##= 14 sin 10° inches, 
and the veloc'ty in feet per second will be 

%7t n x If sin 10° 

v = — . 

60 

= T8o^^sinl0°. 

Employing the equation at the bottom of page 
232 of the text, making F = 4 lbs., and sub- 
stituting the value of v given above, we have 

(jloTtn sin 10°) 2 = 32ir X tt X sin 10° tan 10° x 
4x32|xl4sinl0° . 
5X12 



TO ELEMENTARY MECHANICS. 179 

In f 6 3 [> cos 10° ^5 sin 10° J 



= 118-86. 

The reduction is as follows : 



180 



_ 4/I55I / 1 1 

= 7 X 3141(3 ' 9 U x 0-98481 + 5 x 0-17365 , 

J^-y^ (0- 25 383 +1 -l 5 lT4) 



"21-9912 r 9 



180 i/1351 H . •..„ 
; y -5- x l-405o7. 



" 21-9912 Y 9 

log. 1-40557 0-147852 

log. 1351 3-130655 

log. 9 ar. co. 9*045757 



Dividing by 2, 2*324264 



1 162133 

log. 180 2-255273 

log. 21-9912 a. c. 8 657750 



log. 118 86 2-075055 

12. The tenacity to be overcome will be that in a 
section through the axis of the stone, and 
hence will be 

4.x 12 x 4 x 600 = 115200 lbs. 

The centrifugal force producing rupture will 
be that due to either half of the stone into 
which it is divided by the plane section ; and 
this will equal the mass of one-half multiplied 



180 KEY AND SUPPLEMENT 

by the distance of its centre of gravity from 
the axis of rotation. 

The centre of gravity of a semicircle is -J — 

from the centre of the circle (Ex. 1, p. 154 of 
the text). 

If S he the weight of a cubic foot of the 
stone, the weight of one-half will be 

\nr % x thickness x § 

= k x 4 x | x ^ 

and mass 

If n be the number of revolutions per min- 
ute, then the angular velocity per second will 
be 

n n it 

and the centrifugal force will be 



2 

3 g 



?fe Xf-=llo200; 



J 115200 X 9 X 30 2 x 193 
: r 8 x 2 X (34416) 2 X 6 X<$ 

13786-7 



TO ELEMENTARY MECHANICS. 181 

exeecises. 

Page 235. 

1. It will not. The deviating force will be the 

normal component of the force at the centre. 

2. Because the centrifugal force causes a pressure 

against the side of "the vessel, to balance which 
requires an increase of height of the water. 

The same tendency exists with other sub- 
stances, but in order that any substance shall 
actually be elevated at the outer surface, the 
centrifugal force must overcome the friction 
between its particles. 

3. The diameter of the earth at the equator is 

about 26 miles more than at the poles. If the 
earth were chauged to a sphere, the polar 
diameter would be increased about 13 miles, 
and the equatorial decreased about the same 
amount. The exact change in the dimensions 
involves a knowledge of the volume of ellip_ 
soids. 

4. It does. The angular velocity of the earth 

being constant, the centrifugal force varies as 
the distance from the centre. 

5. It would come to rest on the surface of the hol- 

low. 

6. According to Article 318, if the earth revolved 

in 84 m. 42 T 6 T sec, bodies on the equator would 
weigh nothing ; hence if it revolved in 84 
minutes they would fly off if free, but if held 



1S2 KEY AND SUPPLEMENT 

by cohesion or otherwise, the holding force 
must be overcome before they could fly off. 

7. It would be nearer. 

8. Because the velocity in their orbits is necessarily 

greater in order to balance the attractive force 
of the sun, and the circumference of the orbit 
is less than those more remote. Kepler's law 
is — the squares of the times of the revolutions 
are as the cubes of the mean distances from the 



9. The centrifugal force is not destroyed — it is only 
equilibrated. 

10. Tangentially. 

11. It is proper to say that the centrifugal force — 

and hence the centripetal — is due to the veloc- 
ity of the stone in the sling. Strictly speak- 
ing, the centripetal force has nothing to do 
with the velocity; but indirectly it has, for the 
velocity could not be produced without the 
centripetal force — and it is only in this sense 
that it has something to do with it. 

12. The centrifugal force causes the clothes to 

press against the vertical inside surface of the 

vessel. 

13. They are not affected by the rotation ; but if 

the earth should cease to rotate, the surface 
at the poles would be elevated ; and hence 
cause bodies there to weigh less by being at a 
greater distance from the centre of the earth. 



TO ELEMENTARY MECHANICS. 183 

14. There is. If the moment of the centrifugal 
force in reference to the outer rail as an ori- 
gin exceeds the moment of the weight of the 
car in reference to the same point, they will 
overturn. 

Page 236. — The analysis of this chapter very properly 
belongs to higher analysis ; but we have succeeded, 
by means of curves and special artifices, in bringing it 
within geometrical and algebraic analysis. 

In the language of the calculus, if m be the mass and r] the 
force at a unit's distance, we would have 





cFx 


Multiplying by da 


: gives 


dx d-x 
m 2 = — tjx dx, 


and integrating, 


m fF=-w+io, 


or 






dx 2 

m dJ>=- VX + C ' 


Assuming for the 
x = a, we have 


initial conditions that the velocity is zero, and 




= - r/a 2 + C, 


and we have 


.-. C = r]a\ 




dx 2 V , « 

dt 2 m ' 



which is the square of the velocity, and is the same as equation 
(2), page 233 of the text. From the last equation we have 



-^ 



dt: ■ ^ 



which integrated gives 



184: KEY AND SUPPLEMENT 

a fin 1 x n> 

t = {/ - sin" 1 - + G . 
v ?/ a 

Assuming that t = for * = a, we have 
r 7 



.\t= y — (sin T |3r) 



?/ a 



If a; = 0, we have 



But we also have sin -1 = 7T, for which value, we have 

f ?/ 
which is the same as Equation (4), page 241 of the text. 

Page 243, Art. 328. — Captain Kater used the principle 
of the convertibility of the centres of suspension and 
oscillation for determining the length of a simple 
seconds pendulum, and hence the acceleration due to 
gravity.— Phil. Trans., 1818. 

Let a body, furnished with a movable weight, be 
provided with a point of suspension Ay and another 
point on which it may vibrate, fixed as nearly as can 
be estimated in the centre of oscillation B, and in a 
line with the point of suspension- and the centre of 
gravity. The oscillations of the body must now be 
observed when suspended from A, and also when sus- 
pended from B. If the vibrations in each position 
should not be equal in equal times, they may readily 



TO ELEMENTARY MECHANICS. 185 

be made so by shifting the movable weight. When 
this is done, the distance between the two points A 
and B is the length of the simple equivalent pendu- 
lum. Thus the length L and the corresponding time 
T of vibration will be found uninfluenced by any 
irregularity of density or figure. In these experi- 
ments, after numerous trials of spheres, etc., knife 
edges were preferred as a means of support. At the 
centres of suspension and oscillation there were two 
triangular apertures to admit the knife edges on 
which the body rested while making its oscillations. 

Having thus the means of measuring the length L 
with accuracy, it remains to determine the time T. 
This is effected by comparing the vibrations of the 
body with those of a clock. The time of a single 
vibration or of any small arbitrary number of vibra- 
tions cannot be observed directly, because this would 
require the fraction of a second of time, as shown by 
the clock, to be estimated either by the eye or ear. 
The vibrations of the body may be counted, and here 
there is no fraction to be estimated, but these vibra- 
tions will not probably fit in with the oscillations of 
the clock pendulum, and the differences must be esti- 
mated. This defect is overcome by " the method of 
coincidences." Supposing the time of vibration of 
the clock to be a little less than that of the body, the 
pendulum of the clock will gain on the body, and at 
length at a certain vibration the two will for an in- 
stant coincide. The two pendulums will now be seen 
to separate, and after a time will again approach each 
other, when the same phenomenon will take place. 
If the two pendulums continue to vibrate with per- 



186 KEY AND SUPPLEMENT 

feet uniformity, the number of oscillations of the pen- 
dulum of the clock in this interval will be an integer, 
and the number of oscillations of the body in the 
same interval will be less by one complete oscillation 
than that of the pendulum of the clock. Hence by 
a simple proportion the time of a complete oscillation 
may be found. 

The coincidences were determined in the following 
manner : Certain marks made on the two pendulums 
were observed by a telescope at the lowest point of 
their arcs of vibration. The h'eld of view was limited 
by a diaphragm to a narrow aperture across which the 
marks were seen to pass. At each succeeding vibra- 
tion the clock pendulum follows the other more 
closely, and at last the clock-mark completely covers 
the other during their passage across the field of view 
of the telescope. After a few vibrations it appears 
again preceding the other. The time of disappear- 
ance was generally considered as the time of coinci- 
dence of the vibrations, though in strictness the mean 
of the times of disappearance and reappearance ought 
to have been taken, but the error thus produced is 
very small. (Ehcyc. Met., Figure of the Earth.) 
In the experiments made in Hartan coal-pit in 1854, 
the Astronomer Royal used Kater's method of ob- 
serving the pendulum. {Phil. Trans., 1856.) 

The value of T thus found will require several cor- 
rections. These are called " Reductions." If the 
centre of oscillation does not describe a cycloid, al- 
lowance must be made for the alteration of time de- 
pending on the arc described. This is called " the 
reduction to infinitely small arcs." If the point of 



TO ELEMENTARY MECHANICS. 187 

support be not absolutely fixed, another correction is 
required {Phil. Trans.] 1831). The effect of the 
buoyancy and the resistance of the air must also be 
allowed for. This is the " reduction to a vacuum." 
The length L must also be corrected for changes of 
temperature. 

The time of an oscillation thus corrected enables us 
to find the value of gravity at the place of observa- 
tion. A correction is now required to reduce this re- 
sult to what it w T ould have been at the level of the 
sea. The attraction of the intervening land must be 
allowed for by Dr. Young's rule {Phil. Trans., 1819). 
"We thus obtain the force of gravity at the level of 
the sea, supposing all the land above this level were 
cut off and the sea constrained to keep its present 
level. As the sea would tend in such a case to change 
its level, further corrections are still necessary if we 
wish to reduce the result to the surface of that sphe- 
roid which most nearly represents the earth. (See 
Camb. Phil. Trans., vol. x.) 

There is another use to which the experimental de- 
termination of the length of a simple equivalent pen- 
dulum may be applied. It has been adopted as a 
standard of length on account of being invariable and 
capable at any time of recovery. An Act of Parlia- 
ment (5 Geo. IV.) defines the yard to contain thirty- 
six such parts, of which parts there are 39.1393 in the 
length of the pendulum vibrating seconds of mean 
time in the latitude of London, in vacuo, at the level 
of the sea, at temperature 62° F. The commission- 
ers, however, appointed to consider the mode of re- 
storing the standards of weight and measure which 



188 KEY AND SUPPLEMENT 

were lost by fire in 1831, report that several elements 
of reduction of pendulum experiments are yet doubt- 
ful or erroneous, so tbat the results of a convertible 
pendulum are not so trustworthy as to serve for sup- 
plying a standard for length ; and they recommend 
a material standard, the distance, namely, between 
two marks on a certain bar of metal under given cir- 
cumstances, in preference to any standard derived 
from measuring phenomena in nature. (Report^ 1841.) 

All nations, practically, use this simple mode of 
determining the length of the standard of measure, 
that of placing two marks on a bar, and by a legal 
enactment declaring it to be a certain length. 

For length of seconds pendulum see Mech. Celeste^ 
T. II., pp. 327, 313, 179. 

solutions of examples. 

Page 248. 

1. From the equation on page 213 of the text we 
have 

but in the example t — 4- ; hence 



.1 



9 - 



4?"4x (3.1116) 2 
= 0-827 feet. 
= 9-77 inches. 



TO ELEMENTARY MECHANICS. 189 

2. For this example we have 



9 

. 7 _±g _ 4 x 32| 

* tt 3 (3-1416) 2 

= 13-036 feet. 

3. First find the time, in seconds, of one vibration. 

In one day there are 24. X 60 X 60 = 86,400 
seconds ; hence the time of one vibration will be 

|§|flfl = 0-99976 seconds. 
Let x be the required length, then from the 
equation on page 243 (f being the time for 
length x) we have 

f : t' % : : x : I, 
or 

1 : (0-99976) 2 : : x : 39-1 ; 

•'• :e = (oW = 39 - 1181 + inches; 

hence it must be elongated 

39-1181 - 39-1= 0-0181 + inches. 

Page 249. 

4. From the equation on page 246 of the text we 

have 

45-5 



86354-5 
45-5 



v 



86354-5 
= 11024-3 feet 



20,923,161 feet, 



190 KEY AND SUPPLEMENT 

5. From the equation 

9 
we have 

5t = 15-7080 /A 

= 3-9168 seconds. 

6. From Eq. (2), page 246, we have 



= a/32-0902 x 20,923,161 
= 25911-93 feet per second 
= 4-9 + miles per second. 

7. We will have from Eq. (1), page 246 of the text, 



1=^/1 



= 3-1416 j/ 30,923,161 
K 32-0902 

= 42 m. 17 sec. 

8. From Equations (1) and (4), pages 247 and 248, 
we have 

Fl A /Ekq 

^ = m v ft 

y PEk 



TO ELEMENTARY MECHANICS. 191 



= 3000 |/ : 



1000 x 28000000 x i 
0-453 ft. per sec. 



9. From Eq. (6) we have 
9 

a/pi 



■ 3-1416 



r 29,1. 



1000 x 5 



32±- x 28000000 x J 
== 0-0148 of a second. 

solutions of examples. 
Page 258. 

1. In this example the weight of the fluid is ab- 
stracted, and we have only to consider the 
effect of the pressure of the piston. The area 
of the piston is \nd? — 0-7854 inches ; hence 
the pressure jp per square inch will be 

Z> = 20 ~ 0-7854 = 25-46 + lbs. 

The area of the bottom of the box = 2 x 3 x 
144 == 864 sq. inches. The area of the sides 
= (2 x 1 x 2 + 2 x 1 x 3) 144. = 1440 sq. 
inches. Hence the entire area of the bottom 
and sides will be 2,304 sq. inches; hence the 
pressure will be 

P = 2304 x 25-46 
= 58659-8 lbs. 



192 KEY AND SUPPLEMENT 

2. It will equal the weight of the water; hence 

p =i 8x62 * lbs - , 

= 1-736 lbs. 

3. The area of the base will be 31416 x 16. The 

pressure due to the liquid will equal the weight 
of a cylinder of water whose base is 8 inches in 
diameter and height 10 inches, or 

3-1416 x 16 x 10 aQ1 1Q1Q11 

* x 62| = 18-18 lbs. 

1 i ~o 

The pressure upon the base due to the exter- 
nal pressure of 100 lbs. will be 

?- x 100 = 177-777 lbs. ; 

hence the total pressure will be 

P = 18-208 + 177-777 
= 195-96 lbs. 

4. The upward pressure equals the weight of water 

displaced — or 63 lbs. ; hence the pull on the 
string will be the difference of the upward 
pressure and the weight of the block, or 

63 - 35 = 28 lbs. 

5. We have from Article 86, p. 44 of the text, 

F=Mf; 



TO ELEMENTARY MECHANICS. 193 



. f _F_m _28 x 324 



"Jf~35 y 35 

And, from Eq. (3), p. 12 of the text, 

v = V2fi 



_ ,/ 28 X 64fr x 50 

~ y 35 

= 50-73 ft. per sec. 

Page 259. 

6. We have, for the volume of the block (|) 3 = a 8 L 

feet. Hence the weight = 5 8 x x 180 = 607-5 
lbs. The upward pressure of the water will 
be V- x 62J lbs. = 210-937 lbs., which, sub- 
tracted from the weight, will equal the tension, 
or 396-563 lbs. 

7. Let a equal half the length of the bar, and x the 

distance of the point of attachment from the 
middle of the bar, which point will be in the 
iron part. Let s be the ratio of the weight of 
a given volume of wood to that of an equal 
volume of water. If the weight of the water 
displaced per unit of length of the bar be 
called unity, then will the weight of the 
wooden part of the bar be represented bj sa, 
and of the iron part bv Ssa. The upward 
pressure of the water will be 2a, since it equals 
the volume displaced. The resultant upward 
stress on the wooden part will be a — sa = 
(1 — s)a ; and its moment will be (1 — s)a x 
{a -f x). The resultant downward force of the 



194 KEY AND SUPPLEMENT 

bar between the middle of the beam and point 
of attachment will be (8s — l)x, and the mo- 
ment will be 4 (8* — l)x 2 . The moment of the 
remaining part will be i(Ss — l)(a — xf. 
Hence we have the equation 

i(Ss - \){a - xf = a(1 - s)(a + x) - i(S.y -\)x 2 

Is ± V- 111* 2 + 68* - 
='-f = 16^2" -« 

8. If one end is depressed 3 inches, the other will 
be raised the same amount, hence the differ- 
ence of level will be G inches, and the column 
of water necessary to produce this difference 
of level will be 

6 x 1H = 81 inches. 

ANSWERS TO EXERCISES. 

1. It would not disperse, but would remain in the 

same form. 

2. The gas would till the hollow space. 

3. If the sides of the pail were vertical, it probably 

would ; but even in this case, if it were poured 
in and run down the side, it would require 
a short time to overcome the adhesion on the 
side and permit the entire weight to be exerted 
on the bottom of the vessel. 

4. As much less as the weight of an equal volume 

of air. 

5. Of the same density of air — or the weight must 

be the same as that of an equal volume of air. 



TO ELEMENTARY MECHANICS. 195 

SOLUTIONS OF EXAMPLES. 



Page 270. 



o _ W r SS 2 



^S + 2VS 2 — WS 

substituting the value of s = 1-3077, we have 
12 x 1-3077 x 11 



34 x 11 + (12 - 31)1-3077 



= h 



w 32 32 iA 

A * «7 - 10, 32-25 7 ~ % ' 

3. 5 = -== - 1 . In this case w t = : .*. solving for 

TF we have 

^ ws -8 x 60 48 Q/ln 

W= „ = -q rr = --S- — 240 grains. 

1 — s 1 — -o .2 ° 

4. A stone 5 ft. on each edge = 125 cu. ft. and will 

displace 125 x 62-5 = 7812-5 Bbs. water, x 2-3 
= 17968-75 lbs. = weight of stone. 



w - w 40-32 8 

s = " = Tn ^ = ~ = 1-6. 

w — u\ 40 — 35 o 

c _ w + v 1 s 1 __ 35 + 18 _ 53 __ 

*"«+«! ~ 5+2 ~T~ i ' b12 " 

w _ fe ~ « h ^ = (10-5 -14)19-3 x 10 

1 (*-*i)« (10-5-19 3)14 

— 675-5 K , QQ 

^123^2 = 5 ' 483 - 



196 KEY AND SUPPLEMENT 

(■!?! - *)■% _ (19-3 - 14)10-5 x 10 _ 







^-(., 1 


556-5 
1232 


(19-3 
= 4-518. 


-10.5)14 






9. 


1 _ 

n 


(V 


• + ?V 9 i 


- 27 x 1 + 39-4915 > 
(27 + 39-4915)1- 


c 1-8485 
6321 






= 1 


100 
108-52 


= 1 - -9215 = 0-0785. 








10. 


,-''■ 


4- 0-0013 (b 2 - c) _ 
b\ + b 2 — c 


14 + 0-0013(10 - 

14 + 10-7 


7). 






14-0039 


0-8237. 









17 



ANSWERS TO EXERCISES. 



Page 271. 

1. It will not ; for water being more compressible 

than the solid, will be relatively more dense 
in the air than in a vacuum, and hence when 
in air will force the body upward more than 
when in a vacuum. 

2. Because the smoke is lighter than the surround- 

ing air; but if it be heavier it will fall in the 
air. 

3. See answer to Exercise 1. 

4. This assumes that the weight of the bag — or 

some other cause — causes the bag to sink, and 
if it sinks at all it will go to the bottom of the 
vessel. If now a pressure be exerted upon the 
surface of the liquid, it will cause the bag (or 



TO ELEMENTARY MECHANICS. 197 

gas) to condense more than the liquid, and 
lience it will not rise. Toys have been made 
involving this principle. 

5. Water is more compressible than iron for the 

same pressure, hence it seems possible, theoret- 
ically, for water to be subjected to such a pres- 
sure as to be as dense as iron at the same pres- 
sure. 

6. If both are incompressible, their relative densi- 

ties will be unchanged by pressure, and hence 
the heavier body will sink indefinitely. If the 
body be compressible, it will become relatively 
more dense, and hence there will be no limit. 

7. If the brine be sufficiently " strong " — according 

to popular language — it will float the egg. In 
order that it may float between the top and 
bottom, the brine must be more dense near the 
bottom than at the top. 

8. It will. It is related of Benjamin Franklin, 

that he asked a company of savants why a pail 
of water containing a fish would weigh no more 
than without the fish. Several reasons were 
given, and finally he was appealed to for the 
reason. He thus replied : " Are you sure it 
will weigh no more?" They had been trying 
to explain a false assumption. 

* solutions of examples. 
Page 276. 

f 7? 
1. The equation on page 273 gives us — = -^ = 

tan cp\ 



198 KEY AND SUPPLEMENT 

tan cp = ft =0-3 ; .\ cp = 16° 41' 57". 

2. f=gh\\\ cp. The section of the liquid will be 
a triangle with a base of 3 ft. and a height of 
2 ft., hence tan cp = § ; 

.%/"= | x 32£ = 21f feet per second. 

8. f=g-tan cp. In this case the slope of the free 
surface will be 45°, and tan 45° = 1 ; .'./= g. 

4. Let A be the edge of the vessel. From A con- 
ceive a horizontal line drawn to XE, and mark 
the foot with the letter Z. The volume gen- 
erated by the revolution of the semi -parabola 
is one-half the product of the base and altitude 
(See Mensuration, or works on the Integral 
Calculus). And as the volume of this parabo- 
loid is the unoccupied portion of the cylinder, 
the altitude Z^will be 2 x 3 = 6 inches. AZ 
is 12 inches. From a property of the parabola, 
we have 

EZ : AZ : : AZ : the parameter, 
(orxiyi: y : 2p) ; 

12 2 

.*. the parameter = -- = 24 inches. 
6 

This is known to be twice the subnormal DC; 
hence DC = 12 inches, which, as shown on 
p. 275 of the text, is <j -f- ca 2 , therefore 



12 - 



TO ELEMENTARY MECHANICS. 199 

Let n be the number of turns per minute 

n nit 
sought. Then 27T — - = — — will be the angular 

velocity per second, and we have 

■?i7t\ 2 _ 32J- a 

30/ " 12 ' 

.-. /i = 11-77 turns. 

5. The angular velocity will be |^2^ = n, which is 
the value of go in the value of DC, p. 275 of 

the text ; hence DC = -4r. This is one-half 

the parameter in the equation y 2 = 2j?x; 

hence the equation becomes y 2 = 2 -— x. 

71 



solutions of examples. 
Page 282. 

1. To fulfill this condition we must have 

\dW = idb(9 - A 2 ) 
A 2 = 4-5 
h = 2-121 feet. 

2. The area of each triangle will be -J- x 1-4 x 2-6 

= 1-82. The centre of gravity of the triangle 
whose base will be in the surface will be 
i x 2-6 x sin 56° 35' below the free surface; 
hence the pressure on it will be 

62i x 1-82 x i x 2-6 x sin 56° 35' = 82-28 lbs. ; 
and of the other triangle, 



200 KEY AND SUPPLEMENT 

62i x 1-82 x f x 2-G x sin 5G° 35' = 1G4-57 lbs. 

3. Pressure on concave surface = ^SbJr, "but h = 

27tr; 

.-.jp = I- x G2-5 x 2 x 31416 x = 1767-15 lbs. 

Weight of liquid = m* x h x 6 = 3-1410 x 3 

x 62-5 =589-05 fts. 
Pressure on base = weight of liquid = 589*05 
lbs. 

4. The weight of the liquid = \$m* = { x 62-5 x 

3-1416 x 125 = 32725-0 lbs. 
Normal pressure = 4#7r/' 3 = 4 x 62*5 x 3-1116 
x 125 = 98175 lbs. 

5. Pressure on flood-gate = ■£d7>(Zf 2 2 — J<\) = £ x 

62-5 x 2 x (l3 2 - 10 8 ) = 62-5 x 69 = 43125 ft»s. 

6. Pressure on opposite side == iSl(7i' 2 — A" 2 ) = £ x 

2 x 62-5(7 2 - I 2 ) = 62-5 x 33 = 2062-5, 

and 
4312-5 - 20G2-5 = 2250 fibs. 

solutions of examples. 
Page 287. 

1. By Art. 372, the centre of pressures of rectangle 

is at a distance from the top equal to J the 
height, /. | of 3 = 2 feet. 

2. By Art. 373, we have for the required depth, 

35 1.5 / y + 1Q \ 35 _ _5 . 1? _ 13G3 _ 9 JL 
6 3 \V + ■ 5 / ~~ « " J8 " 13 ~ 234 ~ 234 



TO ELEMENTARY MECHANICS. 901 

= 1 ^- = 5-427 ft.; or -427 ft. = 519 inches 
below the top of the flood-gate. 

3. To resist overturning we have 

Sb 2 h = ^-hi substituting the values given, 
and we have 

_ 125 x 512 __ 64000 
° ~6 xl80 x8~~ 8640 ~ IW *> 



... I = V7-4074 = 2-72 ft. = 2 feet 8| inches. 

4. In this case we have, p. 279 of the text, 31J&A 2 

lbs. for the pressure of the liquid. Its mo- 
ment in reference to the edge of the wall will 
be J x 31JM 8 , and the wall must be capable 
of resisting twice this amount. The moment 
of the wall will be 

ix4x8xl20xfof4 = 5120 lbs. ; 
.-. f x 31J5A 3 = 5120 ; 
.-. h = 6-2 feet. 

5. The pressures are proportional to the areas, and 
" the areas are as (15) 2 -? (1'5) 2 = 100 to 1 ; 

.-. total pressure = 500 x 100 = 50,000 lbs. 

solutions of examples. 

Page 310. 

2(g __ 2n?*h 



1. t- 



msV^g/i 0-62 x x(&)W64$ x 3 
9* 



202 KEY AND SUPPLEMENT 

2 x i x 3 x 2301 4A1 Q „ ' a 

= = = 401-2 sec. = Cm. 41-2 sec. 

0-62V193 

2. Q = f/rc^ V^/A 3 = | x 06-2 x 2(45 x GO) 

= 13618-8 cubic feet. 

3. The equation on p. 309 of the text is x — 

— 2 if. But h = 24 inches, and r = 3 inches, 

•i i 24 4 8 4 

hence we have a? = — ^ = — y 4 . 

To find the area of the orifice, we have on 

p. 309 of the text, 

Tt^c , A 2 1 

ms = ■ — =r^ , when c = — • = - — = - — • ; 
V2gh T GOO 300' 

r == J ft, A = 2 ft. ; 

3.14.1 fi v 1 v 1 
/# , = ? 1*1" > < TIT X _3i n7 _ . Q 00931 ft 

0-62 V04J x 2 

-0 01341 sq. in. 

4. From the equation on p. 303 of the text we have 

25 - 0-025187 /£ = 0-0006769 1 ?~ 

a) 4 (w 2 

((? + 0-141 724£G-) 2 ) 
or 

25 - 32-64235 Q l = 7895-36(<2 2 + 0-0039360 ; 
. ^ 81-076 Q _ J>5_. 
*' V 7928 V 7928' 

.-. Q = 00542 cu. ft. per sec. 
= 195-1 + cu. ft. per hour. 



TO ELEMENTARY MECHANICS.. 203 

answers to exercises. 
Page 310. 

1. The time will be the same for each. 

2. It will not. The time will be less, for the head 

producing the velocity will be equivalent 
to what it would be if for the weight of the 
water an equal weight of mercury be substi- 
tuted. 

3. It will not. The How of the water in this case 

will exceed that of the mercury in the preced- 
ing. 

4. It will. It may be observed that when the pres- 

sure producing the How of aliqnid is the weight 
of the same liquid, the head equals the height 
of the liquid above the orifice. 

5. It will not, but the depth of submergence will 

gradually increase. The block receives an ini- 
tial velocity downward which is being gradu- 
ally lessened as the surface of the liquid de- 
scends. 

6. It will be lowest near the orifice. 

7. It will be greater ; for the acceleration upward 

of the vessel will have the same effect as an 
increased pressure on the surface. 

8. It will be greater ; for the head over the orifice • 

will be greater. 

9. It is not ; a part of the pressure is engaged in 

producing motion of the mass. 



204 KEY AND SUPPLEMENT 

Pages 326-329. — The expression for the pressure (or 
rather the tension) of the air at any height, x, above 
the earth, p. 326, reduces to 

en 

where p Q = 15 lbs., e = 2-71828 + , and H= 26214 

feet, according to Rankine ; 

15 

.*. v = 



(2.71828)2^4 



If as = 20,000,000 feet, about the radius of the earth, 
we have 

p = wr&sr = 2^nW ,bs - per ■* incl1 ' 

near]}-. 

But this expression gives too rapid a diminution 
of the tension, since the effect of gravity is discarded. 
Newton, in the Princijna, gives the following: 

Proposition xxii., B. II. — Let the density of any 
fluid he 'proportional to the compression, and its parts 
attracted downwards hy a gravitation reciprocally 
proportional to the squares of the distana from t he- 
centre. I say, that if the distances be taken in har- 
monic progression, the densities of the fluid at those 
distances irill he in geometrical progression. In ac- 
cordance with this proposition the following table 
was computed in The System of the World by Sir 
Isaac Newton : 



TO ELEMENTARY MECHANICS. 



205 





COMPRESSION OF THE AIR. 




HEIGHT IN MILFS. 


Initial pressure eql 
a column of ava^er 


AL 

33 


EXPANSION OF THE AIR, THE 
INITIAL VOLUME BEING 




FEET HIGH. 




UNITY. 





33 




1 


5 


17.1815 




1.848G 


10 


. 9.6717 




3.4151 


20 


2.852 




11.571 


40 


0.2525 




136.83 


400 


0.(10 17 )1224 




26956 x 10 15 


4.000 


0.(10 105 )465 




73907 x 10' 02 


40,000 


0.(10 192 )1628 




20263 x 10 1 * 9 


400,000 


O.(10 210 )7895 




41798 x 10 207 


4,000,000 


0.(10 2J2 )9878 




33414 x 10 209 


Infinite. 

4 


0.(10 2,2 )6041 




54622 x 10 209 



where (10 17 )1224, implies that there are 17 cyphers 
before 1224 in the decimal, thus 000000000000000- 
001224, and similarly for the others. The following 
inference is drawn : " But from this table it ap- 
pears that the air, in proceeding upwards, is rare- 
fied in such a manner, that a sphere of that air 
which is nearest to the earth, of but one inch in 
diameter, if dilated with that rarefaction which it 
would have at the height of one semi-diameter of the 
earth, would fill all the planetary regions as far as the 
sphere of Saturn, and a great way beyond ; and at 
the height of ten semi-diameters of the earth would 
fill up more space than is contained in the whole 
heavens on this side the fixed stars, according to the 
preceding computation of their distance. And though, 
by reason of the far greater thickness of the atmos- 
pheres of comets, and the great quantity of the circum- 
solar centripetal force, it may happen that the air in 
celestial spaces, and in the tails of comets, is not so 



20G KEY AND SUPPLEMENT 

vastly rarefied, yet from this computation it is plain 
that a very small quantity of air and vapor is abund- 
antly sufficient to produce all the appearances of the 
tails of comets, for that they are indeed of a very 
notable rarity appears from the shining of the stare 
through them." Similar remarks are made in the 
Principia under Pjrop. xli. 

solutions of examples. 
Page 331. 

1. In the first formula on p. 317 of the text, mak- 

ing V t = 2 V and a = 0-002039, we find 

Page 332. 

2. At the surface of the earth the pressure of the 

air will balance a column of water 34 feet 
high. The pressures will be inversely as the 
amount of compression ; hence 

1 : 30 : : 34 : x = 1020 feet of water. 

But the first 34 is due to atmospheric pressure ; 
hence the depth will be 1020 - 34 = i>8<> feet. 

3. From the equation in Prob. 3, p. 323 of the text 

we have 

2y^ 2 V ' J { ;\ =2f f±-= :K'0-U041102 
J 34 + 1000 r 8272 

- 0-322 of an inch. 

4. Assuming that a cubic foot of air weighs 0-08<>7- 

of a pound, 10,000 cubic feet will weigh 807-2 



TO ELEMENTARY MECHANICS. 907 

lbs., and the weight of the gas will be 807-2 x 
0-069 = 55-696 lbs. This subtracted from the 
weight of an equal volume of air will give the 
lifting capacity, or 807-2 - 55-696 ='741-504: 
lbs. 
5. From the last equation of Article 419, p. 317 of 
the text, we have (t Q being 32°) 

F, 1 + a(t 2 -t Q ) 

5 1 + 0-002039(400 - 32) 
~~ 5^5 X 1 + 0-002039(32 -32) 
= 23-868 lbs. 



Why will a wheel with nearly all the matter con- 
centrated in the axle roll down a plane in less time 
than if it be nearly all concentrated in the rim ? 

Because, in descending the plane the entire work 
is done by gravity, and the measure of that work is 
the weight into the vertical descent of the centre of 
the wheel ; and this work is changed into kinetic en- 
ergy in the wheel. When the matter is concentrated 
in the axle, the energy will be ^nv*, where m is the 
mass, and v the final velocity of the centre ; but if it 
be all concentrated in the rim its energy will be 
\mvy + i?n(?'Gj) 2 : where m is the same mass as before, 
Vi the final velocity of the centre, & the final angular 
velocity of the rim, and r the radius of the rim. 
Since the total kinetic energy must be the same in 
both cases, it is evident that v must exceed t\, and 
hence the time of descent in the latter case will ex- 



208 KEY T0 ELEMENTARY MECHANICS. 

eeed that in the former. It is shown in Analytical 
Mechanics, p. 215, that if a given mass be entirely con- 
centrated in the axle, or uniformly distributed as a disc, 
or entirety concentrated in the rim, all having the 
same radius, the times will be as V2 : V^ : V4. 



